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CBSE Class 9th Maths 2023 : 30 Most Important Case Study Questions with Answers; Download PDF
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CBSE Class 9 Maths exam 202223 will have a set of questions based on case studies in the form of MCQs. CBSE Class 9 Maths Question Bank on Case Studies given in this article can be very helpful in understanding the new format of questions.
Each question has five subquestions, each followed by four options and one correct answer. Students can easily download these questions in PDF format and refer to them for exam preparation.
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CBSE Class 9 All Students can also Download here Class 9 Other Study Materials in PDF Format.
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CBSE Class 9th Exam 202425 : Skill Subject Sample Papers and Marking Scheme Released; Download PDF
CBSE has released the 202425 Class 9th skill subject sample papers and marking schemes. These resources help students understand the exam pattern, improve time management, and boost confidence. Download the PDFs from the CBSE official website to enhance your preparation. Practice consistently with these sample papers to excel in your exams and gain practical knowledge in key skill subjects.
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CBSE Class 9 Maths Most Important Case Study Based Questions With Solution
Cbse class 9 mathematics case study questions.
In this post I have provided CBSE Class 9 Maths Case Study Based Questions With Solution. These questions are very important for those students who are preparing for their final class 9 maths exam.
All these questions provided in this article are with solution which will help students for solving the problems. Dear students need to practice all these questions carefully with the help of given solutions.
As you know CBSE Class 9 Maths exam will have a set of cased study based questions in the form of MCQs. CBSE Class 9 Maths Question Bank given in this article can be very helpful in understanding the new format of questions for new session.
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Case studies in class 9 mathematics.
The Central Board of Secondary Education (CBSE) has included case study based questions in the Class 9 Mathematics paper in current session. According to new pattern CBSE Class 9 Mathematics students will have to solve case based questions. This is a departure from the usual theoretical conceptual questions that are asked in Class 9 Maths exam in this year.
Each question provided in this post has five subquestions, each followed by four options and one correct answer. All CBSE Class 9th Maths Students can easily download these questions in PDF form with the help of given download Links and refer for exam preparation.
There is many more free study materials are available at Maths And Physics With Pandey Sir website. For many more books and free study material all of you can visit at this website.
Given Below Are CBSE Class 9th Maths Case Based Questions With Their Respective Download Links.
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myCBSEguide
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 CBSE Class 9 Mathematics...
CBSE Class 9 Mathematics Case Study Questions
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Download the app to get CBSE Sample Papers 202324, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.
If you’re looking for a comprehensive and reliable study resource and case study questions for class 9 CBSE, myCBSEguide is the perfect door to enter. With over 10,000 study notes, solved sample papers and practice questions, it’s got everything you need to ace your exams. Plus, it’s updated regularly to keep you aligned with the latest CBSE syllabus . So why wait? Start your journey to success with myCBSEguide today!
Significance of Mathematics in Class 9
Mathematics is an important subject for students of all ages. It helps students to develop problemsolving and criticalthinking skills, and to think logically and creatively. In addition, mathematics is essential for understanding and using many other subjects, such as science, engineering, and finance.
CBSE Class 9 is an important year for students, as it is the foundation year for the Class 10 board exams. In Class 9, students learn many important concepts in mathematics that will help them to succeed in their board exams and in their future studies. Therefore, it is essential for students to understand and master the concepts taught in Class 9 Mathematics .
Case studies in Class 9 Mathematics
A case study in mathematics is a detailed analysis of a particular mathematical problem or situation. Case studies are often used to examine the relationship between theory and practice, and to explore the connections between different areas of mathematics. Often, a case study will focus on a single problem or situation and will use a variety of methods to examine it. These methods may include algebraic, geometric, and/or statistical analysis.
Example of Case study questions in Class 9 Mathematics
The Central Board of Secondary Education (CBSE) has included case study questions in the Class 9 Mathematics paper. This means that Class 9 Mathematics students will have to solve questions based on reallife scenarios. This is a departure from the usual theoretical questions that are asked in Class 9 Mathematics exams.
The following are some examples of case study questions from Class 9 Mathematics:
Class 9 Mathematics Case study question 1
There is a square park ABCD in the middle of Saket colony in Delhi. Four children Deepak, Ashok, Arjun and Deepa went to play with their balls. The colour of the ball of Ashok, Deepak, Arjun and Deepa are red, blue, yellow and green respectively. All four children roll their ball from centre point O in the direction of XOY, X’OY, X’OY’ and XOY’ . Their balls stopped as shown in the above image.
Answer the following questions:
Answer Key:
Class 9 Mathematics Case study question 2
 Now he told Raju to draw another line CD as in the figure
 The teacher told Ajay to mark ∠ AOD as 2z
 Suraj was told to mark ∠ AOC as 4y
 Clive Made and angle ∠ COE = 60°
 Peter marked ∠ BOE and ∠ BOD as y and x respectively
Now answer the following questions:
 2y + z = 90°
 2y + z = 180°
 4y + 2z = 120°
 (a) 2y + z = 90°
Class 9 Mathematics Case study question 3
 (a) 31.6 m²
 (c) 513.3 m³
 (b) 422.4 m²
Class 9 Mathematics Case study question 4
How to Answer Class 9 Mathematics Case study questions
To crack case study questions, Class 9 Mathematics students need to apply their mathematical knowledge to reallife situations. They should first read the question carefully and identify the key information. They should then identify the relevant mathematical concepts that can be applied to solve the question. Once they have done this, they can start solving the Class 9 Mathematics case study question.
Students need to be careful while solving the Class 9 Mathematics case study questions. They should not make any assumptions and should always check their answers. If they are stuck on a question, they should take a break and come back to it later. With some practice, the Class 9 Mathematics students will be able to crack case study questions with ease.
Class 9 Mathematics Curriculum at Glance
At the secondary level, the curriculum focuses on improving students’ ability to use Mathematics to solve realworld problems and to study the subject as a separate discipline. Students are expected to learn how to solve issues using algebraic approaches and how to apply their understanding of simple trigonometry to height and distance problems. Experimenting with numbers and geometric forms, making hypotheses, and validating them with more observations are all part of Math learning at this level.
The suggested curriculum covers number systems, algebra, geometry, trigonometry, mensuration, statistics, graphing, and coordinate geometry, among other topics. Math should be taught through activities that include the use of concrete materials, models, patterns, charts, photographs, posters, and other visual aids.
CBSE Class 9 Mathematics (Code No. 041)
I  NUMBER SYSTEMS  10 
II  ALGEBRA  20 
III  COORDINATE GEOMETRY  04 
IV  GEOMETRY  27 
V  MENSURATION  13 
VI  STATISTICS & PROBABILITY  06 
Class 9 Mathematics question paper design
The CBSE Class 9 mathematics question paper design is intended to measure students’ grasp of the subject’s fundamental ideas. The paper will put their problemsolving and analytical skills to the test. Class 9 mathematics students are advised to go through the question paper pattern thoroughly before they start preparing for their examinations. This will help them understand the paper better and enable them to score maximum marks. Refer to the given Class 9 Mathematics question paper design.
QUESTION PAPER DESIGN (CLASS 9 MATHEMATICS)
1.  Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers. Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas  43  54 
2.  Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.  19  24 
3.  Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations Present and defend opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria. Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions  18  22 
80  100 
myCBSEguide: Blessing in disguise
Class 9 is an important milestone in a student’s life. It is the last year of high school and the last chance to score well in the CBSE board exams. myCBSEguide is the perfect platform for students to get started on their preparations for Class 9 Mathematics. myCBSEguide provides comprehensive study material for all subjects, including practice questions, sample papers, case study questions and mock tests. It also offers tips and tricks on how to score well in exams. myCBSEguide is the perfect door to enter for class 9 CBSE preparations.
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16 thoughts on “CBSE Class 9 Mathematics Case Study Questions”
This method is not easy for me
aarti and rashika are two classmates. due to exams approaching in some days both decided to study together. during revision hour both find difficulties and they solved each other’s problems. aarti explains simplification of 2+ ?2 by rationalising the denominator and rashika explains 4+ ?2 simplification of (v10?5)(v10+ ?5) by using the identity (a – b)(a+b). based on above information, answer the following questions: 1) what is the rationalising factor of the denominator of 2+ ?2 a) 2?2 b) 2?2 c) 2+ ?2 by rationalising the denominator of aarti got the answer d) a) 4+3?2 b) 3+?2 c) 3?2 4+ ?2 2+ ?2 d) 2?3 the identity applied to solve (?10?5) (v10+ ?5) is a) (a+b)(a – b) = (a – b)² c) (a – b)(a+b) = a² – b² d) (ab)(a+b)=2(a² + b²) ii) b) (a+b)(a – b) = (a + b
MATHS PAAGAL HAI
All questions was easy but search ? hard questions. These questions was not comparable with cbse. It was totally wastage of time.
Where is search ? bar
maths is love
Can I have more questions without downloading the app.
I love math
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I am Riddhi Shrivastava… These questions was very good.. That’s it.. ..
For challenging Mathematics Case Study Questions, seeking a writing elite service can significantly aid your research. These services provide expert guidance, ensuring your case study is wellresearched, accurately analyzed, and professionally written. With their assistance, you can tackle complex mathematical problems with confidence, leading to highquality academic work that meets rigorous standards.
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Case Study Questions for Class 9 Maths
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Are you preparing for your Class 9 Maths board exams and looking for an effective study resource? Well, you’re in luck! In this article, we will provide you with a collection of Case Study Questions for Class 9 Maths specifically designed to help you excel in your exams. These questions are carefully curated to cover various mathematical concepts and problemsolving techniques. So, let’s dive in and explore these valuable resources that will enhance your preparation and boost your confidence.
Join our Telegram Channel, there you will get various ebooks for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.
CBSE Class 9 Maths Board Exam will have a set of questions based on case studies in the form of MCQs. The CBSE Class 9 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.
If you want to want to prepare all the tough, tricky & difficult questions for your upcoming exams, this is where you should hang out. CBSE Case Study Questions for Class 9 will provide you with detailed, latest, comprehensive & confidenceinspiring solutions to the maximum number of Case Study Questions covering all the topics from your NCERT Text Books !
Table of Contents
CBSE Class 9th – MATHS: Chapterwise Case Study Question & Solution
Case study questions are a form of examination where students are presented with reallife scenarios that require the application of mathematical concepts to arrive at a solution. These questions are designed to assess students’ problemsolving abilities, critical thinking skills, and understanding of mathematical concepts in practical contexts.
Chapterwise Case Study Questions for Class 9 Maths
Case study questions play a crucial role in the field of mathematics education. They provide students with an opportunity to apply theoretical knowledge to realworld situations, thereby enhancing their comprehension of mathematical concepts. By engaging with case study questions, students develop the ability to analyze complex problems, make connections between different mathematical concepts, and formulate effective problemsolving strategies.
 Case Study Questions for Chapter 1 Number System
 Case Study Questions for Chapter 2 Polynomials
 Case Study Questions for Chapter 3 Coordinate Geometry
 Case Study Questions for Chapter 4 Linear Equations in Two Variables
 Case Study Questions for Chapter 5 Introduction to Euclid’s Geometry
 Case Study Questions for Chapter 6 Lines and Angles
 Case Study Questions for Chapter 7 Triangles
 Case Study Questions for Chapter 8 Quadilaterals
 Case Study Questions for Chapter 9 Areas of Parallelograms and Triangles
 Case Study Questions for Chapter 10 Circles
 Case Study Questions for Chapter 11 Constructions
 Case Study Questions for Chapter 12 Heron’s Formula
 Case Study Questions for Chapter 13 Surface Area and Volumes
 Case Study Questions for Chapter 14 Statistics
 Case Study Questions for Chapter 15 Probability
The above Case studies for Class 9 Mathematics will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Maths Case Studies have been developed by experienced teachers of schools.studyrate.in for benefit of Class 10 students.
 Class 9 Science Case Study Questions
 Class 9 Social Science Case Study Questions
How to Approach Case Study Questions
When tackling case study questions, it is essential to adopt a systematic approach. Here are some steps to help you approach and solve these types of questions effectively:
 Read the case study carefully: Understand the given scenario and identify the key information.
 Identify the mathematical concepts involved: Determine the relevant mathematical concepts and formulas applicable to the problem.
 Formulate a plan: Devise a plan or strategy to solve the problem based on the given information and mathematical concepts.
 Solve the problem step by step: Apply the chosen approach and perform calculations or manipulations to arrive at the solution.
 Verify and interpret the results: Ensure the solution aligns with the initial problem and interpret the findings in the context of the case study.
Tips for Solving Case Study Questions
Here are some valuable tips to help you effectively solve case study questions:
 Read the question thoroughly and underline or highlight important information.
 Break down the problem into smaller, manageable parts.
 Visualize the problem using diagrams or charts if applicable.
 Use appropriate mathematical formulas and concepts to solve the problem.
 Show all the steps of your calculations to ensure clarity.
 Check your final answer and review the solution for accuracy and relevance to the case study.
Benefits of Practicing Case Study Questions
Practicing case study questions offers several benefits that can significantly contribute to your mathematical proficiency:
 Enhances critical thinking skills
 Improves problemsolving abilities
 Deepens understanding of mathematical concepts
 Develops analytical reasoning
 Prepares you for reallife applications of mathematics
 Boosts confidence in approaching complex mathematical problems
Case study questions offer a unique opportunity to apply mathematical knowledge in practical scenarios. By practicing these questions, you can enhance your problemsolving abilities, develop a deeper understanding of mathematical concepts, and boost your confidence for the Class 9 Maths board exams. Remember to approach each question systematically, apply the relevant concepts, and review your solutions for accuracy. Access the PDF resource provided to access a wealth of case study questions and further elevate your preparation.
Q1: Can case study questions help me score better in my Class 9 Maths exams?
Yes, practicing case study questions can significantly improve your problemsolving skills and boost your performance in exams. These questions offer a practical approach to understanding mathematical concepts and their reallife applications.
Q2: Are the case study questions in the PDF resource relevant to the Class 9 Maths syllabus?
Absolutely! The PDF resource contains case study questions that align with the Class 9 Maths syllabus. They cover various topics and concepts included in the curriculum, ensuring comprehensive preparation.
Q3: Are the solutions provided for the case study questions in the PDF resource?
Yes, the PDF resource includes solutions for each case study question. You can refer to these solutions to validate your answers and gain a better understanding of the problemsolving process.
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Important Questions Class 9 Maths Chapter 2
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Important Questions Class 9 Mathematics Chapter 2 Polynomials.
Mathematics Chapter 2 of Class 9 is about Polynomials. Polynomial consists of two terms, namely Poly (meaning “many”) and Nominal (meaning “terms.”). A polynomial is explained as an expression which is composed of variables, constants and exponents that are combined using mathematical operations like addition, subtraction, multiplication and division (No division operation by a variable). Based on the number of terms present in the expression, it is classified as monomial, binomial, and trinomial.
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Important Questions Class 9 Mathematics Chapter 2 – With Solutions
Our inhouse Mathematics faculty experts have collated a complete list of Important Questions Class 9 Mathematics Chapter 2 by referring to various sources. The subject experts have meticulously prepared illustration for individual questions that will enable students to comprehend the notions used in each question. Furthermore, the questions are selected in a way that would cover all the topics. So by practising from our question bank, students will be able to revise the chapter and understand their strong and weak topics. And enhance their preparation by further concentrating on weaker sections of the chapter.
Given below are a few of the questions and answers from our question bank of Important Questions Class 9 Mathematics Chapter 2:
Question 1: Calculate the value of 9x² + 4y² if xy = 6 and 3x + 2y = 12.
Answer 1: Consider the equation 3x + 2y = 12
Now, square both sides:
(3x + 2y)² = 12²
=> 9x² + 12xy + 4y² = 144
=>9x² + 4y² = 144 – 12xy
From the questions, xy = 6
9x² + 4y² = 144 – 72
Thus, the value of 9x² + 4y² = 72
Question 2:Evaluate the following using suitable identity
Answer 2: We can write 102 as 100+2
Using identity,(x+y) ³ = x ³ +y ³ +3xy(x+y)
(100+2) ³ =(100) ³ +2 ³ +(3×100×2)(100+2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
Question 3:Without any actual division, prove that the following 2x⁴
– 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.
[Hint: Factorise x² – 3x + 2]
Answer 3: x²3x+2
x(x2)1(x2)
Therefore,(x2)(x1)are the factors.
Considering (x2),
Then, p(x) becomes,
p(x)=2x⁴5x³+2x²x+2
p(2)=2(2)⁴5(2)³+2(2)²2+2
Therefore, (x2) is a factor.
Considering (x1),
p(1)=2(1)⁴5(1)³+2(1)²1+2
Therefore, (x1) is a factor.
Question 4: Using the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case
(i) p(x) = 2x³+x²–2x–1, g(x) = x+1
Answer 4 :p(x) = 2x³+x²–2x–1, g(x) = x+1
∴Zero of g(x) is 1.
p(−1) = 2(−1)³+(−1)²–2(−1)–1
∴By the given factor theorem, g(x) is a factor of p(x).
Question 5: Obtain an example of a monomial and a binomial having degrees of 82 and 99, respectively.
Answer 5: An example of a monomial having a degree of 82 = x⁸²
An example of a binomial having a required degree of 99 = x⁹⁹ + 7
Question 6: If the two x – 2 and x – ½ are the given factors of px ²
+ 5 x + r , show that p = r .
Answer 6: Given, f(x) = px²+5x+r and factors are x2, x – ½
Substituting x = 2 in place of the equation, we get
f(x) = px²+5x+r
f(2) = p(2)²+5(2)+r=0
= 4p + 10 + r = 0 … eq.(i)
Substituting x = ½ in place of the equation, we get,
f( ½ ) = p( ½ )² + 5( ½ ) + r =0
= p/4 + 5/2 + r = 0
= p + 10 + 4r = 0 … eq(ii)
On solving eq(i) and eq(ii),
4p + r = – 10 and p + 4r = – 10
the RHS of both equations are the same,
4p + r = p + 4r
Hence Proved.
Question 7: Identify constant, linear, quadratic, cubic and quartic polynomials from the following.
(i) – 7 + x
(iii) – ? ³
(iv) 1 – y – ? ³
(v) x – ? ³ + ?⁴
(vi) 1 + x + ?²
Answer 7: (i) – 7 + x
The degree of – 7 + x is 1.
Hence, it is a linear polynomial.
The degree of 6y is 1.
Therefore, it is a linear polynomial.
We know that the degree of – ? ³ is 3.
Therefore, it is a cubic polynomial.
We know that the degree of 1 – y – ? ³ is 3.
We know that the degree of x – ? ³ + ?⁴ is 4.
Therefore, it is a quartic polynomial.
We know that the degree of 1 + x + ?² is 2.
Therefore, it is a quadratic polynomial.
We know that the degree of 6?² is 2.
We know that 13 is a constant.
Therefore, it is a constant polynomial.
We know that the degree of –p is 1.
Question 8: Observe the value of the polynomial 5x – 4x² + 3 at x = 2 and x = –1.
Answer 8 : Let the polynomial be f(x) = 5x – 4x² + 3
Now, for x = 2,
f(2) = 5(2) – 4(2)² + 3
=> f(2) = 10 – 16 + 3 = –3
Or, the value of the polynomial 5x – 4x² + 3 at x = 2 is 3.
Similarly, for x = –1,
f(–1) = 5(–1) – 4(–1)² + 3
=> f(–1) = –5 –4 + 3 = 6
The value of the polynomial 5x – 4x² + 3 at x = 1 is 6.
Question 9:Expanding each of the following, using all the suitable identities:
(i) (x+2y+4z)²
(ii) (2x−y+z)²
(iii) (−2x+3y+2z)²
(iv) (3a –7b–c)²
(v) (–2x+5y–3z)²
Answer 9: (i) (x+2y+4z)²
Using identity, (x+y+z)² = x²+²+z²+2xy+2yz+2zx
Here, x = x
(x+2y+4z)² = x²+(2y)²+(4z)²+(2×x×2y)+(2×2y×4z)+(2×4z×x)
= x²+4y²+16z²+4xy+16yz+8xz
Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx
Here, x = 2x
(2x−y+z)² = (2x)²+(−y)²+z²+(2×2x×−y)+(2×−y×z)+(2×z×2x)
= 4x²+y²+z²–4xy–2yz+4xz
Here, x = −2x
(−2x+3y+2z)² = (−2x)²+(3y)²+(2z)²+(2×−2x×3y)+(2×3y×2z)+(2×2z×−2x)
= 4x²+9y²+4z²–12xy+12yz–8xz
Using identity (x+y+z)²= x²+y²+z²+2xy+2yz+2zx
Here, x = 3a
(3a –7b– c)² = (3a)²+(– 7b)²+(– c)²+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)
= 9a² + 49b² + c²– 42ab+14bc–6ca
Here, x = –2x
(–2x+5y–3z)² = (–2x)² + (5y)² + (–3z)² + (2 × –2x × 5y) + (2 × 5y× – 3z)+(2×–3z ×–2x)
= 4x²+25y² +9z²– 20xy–30yz+12zx
Question 10: If the polynomials az³ + 4z² + 3z – 4 and z³ – 4z + leave the same remainder when divided by z – 3, find the value of a.
Answer 10: Zero of the polynomial,
Hence, zero of g(z) = – 2a
Let p(z) = az³+4z²+3z4
Now, substituting the given value of z = 3 in p(z), we get,
p(3) = a (3)³ + 4 (3)² + 3 (3) – 4
⇒p(3) = 27a+36+94
⇒p(3) = 27a+41
Let h(z) = z³4z+a
Now, by substituting the value of z = 3 in h(z), we get,
h(3) = (3)³4(3)+a
⇒h(3) = 2712+a
⇒h(3) = 15+a
As per the question,
The two polynomials, p(z) and h(z), leave the same remainder when divided by z3
So, h(3)=p(3)
⇒15+a = 27a+41
⇒1541 = 27a – a
Question 11: Compute the perimeter of a rectangle whose area is 25x² – 35x + 12.
Answer 11: A rea of rectangle = 25x² – 35x + 12
We know the area of a rectangle = length × breadth
So, by factoring 25x² – 35x + 12, the length and breadth can be obtained.
25x² – 35x + 12 = 25x² – 15x – 20x + 12
=> 25x² – 35x + 12 = 5x(5x – 3) – 4(5x – 3)
=> 25x² – 35x + 12 = (5x – 3)(5x – 4)
Thus, the length and breadth of a rectangle are (5x – 3)(5x – 4).
So, the perimeter = 2(length + breadth)
Therefore, the perimeter of the given rectangle = 2[(5x – 3)+(5x – 4)]
= 2(5x – 3 + 5x – 4)
= 2(10x – 7)
= 20x – 14
Hence, the perimeter of the rectangle = 20x – 14
Question 12: 2x²+y²+²–2√2xy+4√2yz–8xz
Answer 12: Using identity, (x +y+z)² = x²+y²+z²+2xy+2yz+2zx
We can say that, x²+²+²+2xy+2yz+2zx = (x+y+z)²
2x²+y²+8z²–2√2xy+4√2yz–8xz
= (√2x)²+(y)²+(2√2z)²+(2×√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)
= (−√2x+y+2√2z)²
= (−√2x+y+2√2z)(−√2x+y+2√2z)
Question 13: If ? + 2? is a factor of ? ⁵ – 4?²?³ + 2? + 2? + 3, find a.
Answer 13: According to the question,
Let p(x) = x ⁵ – 4a²x³ + 2x + 2a + 3 and g(x) = x + 2a
⟹ x + 2a = 0
Hence, zero of g(x) = – 2a
As per the factor theorem,
If g(x) is a factor of p(x), then p( – 2a) = 0
So, substituting the value of x in p(x), we get,
p ( – 2a) = ( – 2a) ⁵ – 4a²( – 2a)³ + 2( – 2a) + 2a + 3 = 0
⟹ – 32a ⁵ + 32a ⁵ – 2a + 3 = 0
⟹ – 2a = – 3
Question 14: Find the value of x³+ y ³ + z ³ – 3xyz if x² + y² + z² = 83 and x + y + z = 1
Answer 14: Consider the equation x + y + z = 15
From algebraic identities, we know that (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(x + y + z)² = x² + y² + z² + 2(xy + yz + xz)
From the question, x² + y² + z²= 83 and x + y + z = 15
152 = 83 + 2(xy + yz + xz)
=> 225 – 83 = 2(xy + yz + xz)
Or, xy + yz + xz = 142/2 = 71
Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca),
x ³ + y ³ + z ³ – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz))
x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71
So, x ³ + y ³ + z ³ – 3xyz = 15(83 – 71)
=> x ³ + y ³ + z ³ – 3xyz = 15 × 12
Or, x ³ + y ³ + z ³ – 3xyz = 180
Question 15:Verify that:
(i) x³+y³ = (x+y)(x²–xy+y²)
(ii) x³–y³ = (x–y)(x²+xy+y²)
Answer 15: (i) x³+y³ = (x+y)(x²–xy+y²)
We know that (x+y)³= x³+y³+3xy(x+y)
⇒ x³+y³ = (x+y)³–3xy(x+y)
⇒ x³+y³ = (x+y)[(x+y)²–3xy]
Taking (x+y) common ⇒ x³+y³ = (x+y)[(x²+y²+2xy)–3xy]
⇒ x³+y³ = (x+y)(x²+y²–xy)
(ii) x³–y³ = (x–y)(x²+xy+y²)
We know that (x–y)³ = x³–y³–3xy(x–y)
⇒ x³−y³ = (x–y)³+3xy(x–y)
⇒ x³−y³ = (x–y)[(x–y)²+3xy]
Taking (x+y) common ⇒ x³−y³ = (x–y)[(x²+y²–2xy)+3xy]
⇒ x³+y³ = (x–y)(x²+y²+xy)
Question 16: For what value of m is ?³ – 2??² + 16 divisible by x + 2?
Answer 16: According to the question,
Let p(x) = x³ – 2mx² + 16, and g(x) = x + 2
⟹ x + 2 = 0
Hence, zero of g(x) = – 2
if p(x) is divisible by g(x), then the remainder of p(−2) should be zero.
Thus, substituting the value of x in p(x), we obtain,
p( – 2) = 0
⟹ ( – 2)³ – 2m( – 2)² + 16 = 0
⟹ 0 – 8 – 8m + 16 = 0
Question 17:If a + b + c = 15 and a² + b² + c² = 83, find the value of a³ + b³ + c³ – 3abc.
Answer 17: We know that,
a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca) ….(i)
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca ….(ii)
Given, a + b + c = 15 and a² + b² + c² = 83
From (ii), we have
152 = 83 + 2(ab + bc + ca)
⇒ 225 – 83 = 2(ab + bc + ca)
⇒ 142/2 = ab + bc + ca
⇒ ab + bc + ca = 71
Now, (i) can be written as
a³ + b³ + c³ – 3abc = (a + b + c)[(a² + b² + c² ) – (ab + bc + ca)]
a³ + b³+ c³ – 3abc = 15 × [83 – 71] = 15 × 12 = 180.
Question 18: Factorise: 27x³+y³+z³–9xyz
Answer 18: The expression27x³+y³+z³–9xyz can be written as (3x)³+y³+z³–3(3x)(y)(z)
27x³+y³+z³–9xyz = (3x)³+y³+z³–3(3x)(y)(z)
We know that x³+y³+³–3xyz = (x+y+z)(x²+y²+z²–xy –yz–zx)
= (3x+y+z)[(3x)²+y²+z²–3xy–yz–3xz]
= (3x+y+z)(9x²+y²+²–3xy–yz–3xz)
Question 19: If (x – 1/x) = 4, then evaluate (x² + 1/x²) and (x⁴ + 1/x⁴).
Answer 19: Given, (x – 1/x) = 4
Squaring both sides, we get,
(x – 1/x)² = 16
⇒ x² – 2.x.1/x + 1/x² = 16
⇒ x² – 2 + 1/x² = 16
⇒ x² + 1/x² = 16 + 2 = 18
∴ (x² + 1/x²) = 18 ….(i)
Again, squaring both sides of (i), we get
(x² + 1/x²)² = 324
⇒ x⁴ + 2.x².1/x² + 1/x⁴= 324
⇒ x⁴ + 2 + 1/x⁴ = 324
⇒ x⁴ + 1/x⁴ = 324 – 2 = 322
∴ (x⁴ + 1/x⁴) = 322.
Question 20: Factorise
Answer 20: The expression 64m³–343n³ can be written as (4m)³–(7n)³
64m³–343n³ =(4m)³–(7n)³
We know that x³–y³ = (x–y)(x²+xy+y²)
64m³–343n³ = (4m)³–(7n)³
= (4m7n)[(4m)²+(4m)(7n)+(7n)²]
= (4m7n)(16m²+28mn+49n² )
Question 21: Find out the values of a and b so that (2x³ + ax² + x + b) has (x + 2) and (2x – 1) as factors.
Answer 21: Let p(x) = 2x³ + ax² + x + b. Then, p( –2) = and p(½) = 0.
p(2) = 2(2)³ + a(2)² + 2 + b = 0
⇒ –16 + 4a – 2 + b = 0 ⇒ 4a + b = 18 ….(i)
p(½) = 2(½)³ + a(½)² + (½) + b = 0
⇒ a + 4b = –3 ….(ii)
On solving (i) and (ii), we get a = 5 and b = –2.
Hence, a = 5 and b = –2.
Question 22: Explain that p – 1 is a factor of p¹⁰ – 1 and p¹¹ – 1.
Answer 22: According to the question,
Let h(p) = ?¹⁰ − 1,and g(p) = ? – 1
zero of g(p) ⇒ g(p) = 0
Therefore, zero of g(x) = 1
We know that,
According to the factor theorem, if g(p) is a factor of h(p), then h(1) should be zero
h(1) = (1) ¹⁰ − 1 = 1 − 1 = 0
⟹ g (p) is a factor of h(p).
Here, we have h(p) = ?¹¹ − 1, g (p) = ? – 1
Putting g (p) = 0 ⟹ ? − 1 = 0 ⟹ ? = 1
As per the factor theorem, if g (p) is a factor of h(p),
Then h(1) = 0
⟹ (1) ¹¹ – 1 = 0
Hence, g(p) = ? – 1 is the factor of h(p) = ? ¹⁰ – 1
Question 23: Examine whether (7 + 3x) is a factor of (3×3 + 7x).
Answer 23: Let p(x) = 3×3 + 7x and g(x) = 7 + 3x. Now g(x) = 0 ⇒ x = –7/3.
By the remainder theorem, p(x) is divided by g(x), and then the remainder is p(–7/3).
Now, p(–7/3) = 3(–7/3)3 + 7(–7/3) = –490/9 ≠ 0.
∴ g(x) is not a factor of p(x).
Question 24:Prove that:
x³+y³+z³–3xyz = (1/2) (x+y+z)[(x–y)²+(y–z)²+(z–x)²]
Answer 24: We know that,
x³+y³+z³−3xyz = (x+y+z)(x²+y²+z²–xy–yz–xz)
⇒ x³+y³+z³–3xyz = (1/2)(x+y+z)[2(x²+y²+z²–xy–yz–xz)]
= (1/2)(x+y+z)(2×2+2y²+²–2xy–2yz–2xz)
= (1/2)(x+y+z)[(x²+y²−2xy)+(y²+z²–2yz)+(x²+z²–2xz)]
= (1/2)(x+y+z)[(x–y)²+(y–z)²+(z–x)²]
Question 25: Find out which of the following polynomials has x – 2 a factor:
(i) 3?² + 6?−24.
(ii) 4?² + ?−2.
Answer 25: (i) According to the question,
Let p(x) =3?² + 6?−24 and g(x) = x – 2
g(x) = x – 2
zero of g(x) ⇒ g(x) = 0
Hence, zero of g(x) = 2
Thus, substituting the value of x in p(x), we get,
p(2) = 3(2)² + 6 (2) – 24
= 12 + 12 – 24
the remainder = zero,
We can derive that,
g(x) = x – 2 is factor of p(x) = 3?² + 6?−24
(ii) According to the question,
Let p(x) = 4?² + ?−2 and g(x) = x – 2
p(2) = 4(2)² + 2−2
Since the remainder ≠ zero,
We can say that,
g(x) = x – 2 is not a factor of p(x) = 4?² + ?−2
Question 26: Factorise x² + 1/x² + 2 – 2x – 2/x.
Answer 26 : x² + 1/x² + 2 – 2x – 2/x = (x² + 1/x² + 2) – 2(x + 1/x)
= (x + 1/x)² – 2(x + 1/x)
= (x + 1/x)(x + 1/x – 2).
Question 27: Factorise
8a³+b³+12a²b+6ab²
Answer 27: The expression, 8a³+b³+12a²b+6ab² can be written as (2a)³+b³+3(2a)²b+3(2a)(b)²
8a³+b³+12a²b+6ab² = (2a)³+b³+3(2a)²b+3(2a)(b)²
= (2a+b)(2a+b)(2a+b)
Here, the identity, (x +y)³ = x³+y³+3xy(x+y) is used.
Question 28: By Remainder Theorem, find out the remainder when p(x) is divided by g(x), where
(i) p(?) = ?³ – 2?² – 4? – 1, g(?) = ? + 1
(ii) p(?) = ?³ – 3?² + 4? + 50, g(?) = ? – 3
(iii) p(?) = 4?³ – 12?² + 14? – 3, g(?) = 2? – 1
(iv) p(?) = ?³ – 6?² + 2? – 4, g(?) = 1 – 3/2 ?
Answer 28: (i) Given p(x) = ?³ – 2?² – 4? – 1 and g(x) = x + 1
Here zero of g(x) = – 1
By applying the remainder theorem
P(x) divided by g(x) = p( – 1)
P ( – 1) = ( – 1)³ – 2 ( – 1)² – 4 ( – 1) – 1 = 0
Therefore, the remainder = 0
(ii) given p(?) = ?³ – 3?² + 4? + 50, g(?) = ? – 3
Here zero of g(x) = 3
By applying the remainder theorem p(x) divided by g(x) = p(3)
p(3) = 3³ – 3 × (3)² + 4 × 3 + 50 = 62
Therefore, the remainder = 62
(iii) p(x) = 4x³ – 12x² + 14x – 3, g(x) = 2x – 1
Here zero of g(x) = ½
By applying the remainder theorem p(x) divided by g(x) = p (½)
P( ½ ) = 4( ½ )³ – 12( ½ )² + 14 ( ½ ) – 3
= 4/8 – 12/4 + 14/2 – 3
= ½ + 1
= 3/2
Hence, the remainder = 3/2
so, zero of g(x) = 2/3
By applying the remainder theorem p(x) divided by g(x) = p(2/3)
p(2/3) = (2/3)³ – 6(2/3)² + 2(2/3) – 4
Therefore, the remainder = – 136/27
Question 29:Factorise x² – 1 – 2a – a².
Answer 29: x² – 1 – 2a – a² = x² – (1 + 2a + a²)
= x² – (1 + a)²
= [x – (1 – a)][x + 1 + a]
= (x – 1 – a)(x + 1 + a)
∴ x² – 1 – 2a – a² = (x – 1 – a)(x + 1 + a).
Question 30:Evaluate the following using suitable identity
Answer 30: We can write 99 as 1000–2
Using identity,(x–y)³ = x³ –y³ –3xy(x–y)
(998)³ =(1000–2)³
=(1000)³ –2³ –(3×1000×2)(1000–2)
= 1000000000–8–6000(1000– 2)
= 1000000000–8 6000000+12000
= 994011992
Question 31: Find the zeroes of the polynomial:
p(?)= (? –2)² −(? + 2)²
Answer 31: p(x) = (? –2)² −(? + 2)²
Zero of the polynomial p(x) = 0
Hence, we get,
⇒ (x–2)² −(x + 2)² = 0
Expanding using the identity, a² – b² = (a – b) (a + b)
⇒ (x – 2 + x + 2) (x – 2 –x – 2) = 0
⇒ 2x ( – 4) = 0
Therefore, the zero of the polynomial = 0
Benefits Of Solving Important Questions Class 9 Mathematics Chapter 2
Consistently solving questions is a vital element of mastering Mathematics. By solving Mathematics Class 9 Chapter 2 important questions, students can get a further understanding of the polynomials chapter.
A few other advantages of solving Important Questions Class 9 Mathematics Chapter 2 are:
 Class 9 Mathematics Chapter 2 important questions provide details about the types of questions that may be expected in the exams, which increases their confidence in achieving a high grade. .
 Studying something once may help you understand the concepts, but through revisions and guided practice make them aware of their mistakes and help to get the best results. Your chances of getting good grades on exams increase as you solve all the questions from our question bank of Important Questions Class 9 Mathematics Chapter 2.
 The questions and answers provided are based on the NCERT books and the latest CBSE syllabus and examination guidelines. So the students can have complete faith and trust in Extramarks because CBSE itself recommends NCERT books and students can rely on the authenticity of our solutions.
 By solving our Chapter 2 Class 9 Mathematics important questions, students will understand the question paper pattern.. Practising questions identical to the exam questions would help the students become confident in solving advanced level questions with ease and get 100% results in their exams.
Extramarks leaves no stone unturned to give the best learning material to students while combining fun and learning activities through its own study materials to enhance their learning experience. It provides comprehensive learning solutions for students from Class 1 to Class 12. Our website has abundant resources, along with important questions and solutions. Students can click on the links given below to access some of these resources:
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 CBSE extra questions
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Q.1 By actual division, find the quotient and the remainder when x 5 + 1 is divided by x 1
Marks: 3 Ans
x 4 + x 3 + x 2 + x + 1 x 1 x 5 + 1 x 5 x 4 + x 4 + 1 x 4 x 3 + x 3 + 1 x 3 x 2 + ¯ x 2 + 1 x 2 x + x + 1 x 1 + 2 ¯ ¯ ¯ ¯ Quotient : x 4 + x 3 + x 2 + x + 1 Remainder : 2
Q.2 Find the value of k if x 5 is a factor of kx 2 + 3x + 7.
Marks: 2 Ans
Zero of x 5 is 5 as x 5 = 0 gives x = 5 . p(x) = kx 2 + 3 x + 7 p ( 5 ) = 0 25 k + 15 + 7 = 0 25 k + 22 = 0 k = 22 25
Q.3 If x + y + z = 6 and xy + yz + zx = 11, then find the value of x 2 + y 2 + z 2 .
Given : x + y + z = 6 and xy + yz + zx = 11 Squaring both sides , we get x + y + z 2 = 6 2 x 2 + y 2 + z 2 + 2 xy + 2 yz + 2 zx = 36 x 2 + y 2 + z 2 + 2 xy + yz + zx = 36 x 2 + y 2 + z 2 + 2 11 = 36 Since xy + yz + zx = 11 x 2 + y 2 + z 2 + 22 = 36 x 2 + y 2 + z 2 = 36 22 = 14
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Faqs (frequently asked questions), 1. what are the four types of polynomials.
The 4 types of polynomials are zero polynomial, linear polynomial, quadratic polynomial, and cubic polynomial.
2. Where can I get important questions for Class 9 Mathematics Chapter 2 online?
On the Extramarks website, you can find all of the important questions for Class 9 Mathematics Chapter 2, along with their answers. On the website, you can also find important questions and NCERT solutions for all classes from 1 to 12.
3. What are the important chapters in Class 9 Mathematics?
The NCERT Mathematics book has 15 chapters. Each chapter is equally important when it comes to learning the fundamentals and taking the test. Additionally, because CBSE does not specify the distribution of marks for each chapter, students are advised to fully study all chapters. Each and every chapter must be completely understood to acquire a good grade in exams.
All the fifteen chapters of CBSE Class 9 Mathematics syllabus are given below:
 Chapter – Number Systems
 Chapter – Polynomials
 Chapter – Coordinate Geometry
 Chapter – Linear Equations In Two Variables
 Chapter – Introduction To Euclid’s Geometry
 Chapter – Lines And Angles
 Chapter – Triangles
 Chapter – Quadrilaterals
 Chapter – Areas Of Parallelograms And Triangles
 Chapter – Circles
 Chapter – Constructions
 Chapter – Heron’s Formula
 Chapter – Surface Areas And Volumes
 Chapter – Statistics
 Chapter – Probability
CBSE Related Links
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CBSE Important Questions for Class 9 Maths are available in Printable format for Free Download.Here you may find NCERT Important Questions and Extra Questions for Class 9 Mathematics chapter wise with answers also. These questions will act as chapter wise test papers for Class 9 Mathematics. These Important Questions for Class 9 Mathematics are as per latest NCERT and CBSE Pattern syllabus and assure great success in achieving high score in Board Examinations
Total Papers :
Class 9 Maths Marks Distribution  

Units  Marks 
Number Systems  08 
Algebra  17 
Coordinate Geometry  04 
Geometry  28 
Mensuration  13 
Statistics & Probability  10 
Total  80 
Internal Assessment  20 
Grand Total  100 
Maths Topics to be covered for Class 9
 Real Numbers
 Polynomials
 Coordinate Geometry
 Linear Equations in Two Variables
 Introduction to Euclid's Geometry
 Lines and Angles
 Heron’s Formula
 Probability
 Quadrilaterals
 Area of Parallelogram and Triangles
 Constructions
 Surface Areas and Volumes
Structure of CBSE Maths Sample Paper for Class 9 is
Type of Question  Marks per Question  Total No. of Questions  Total Marks 

Objective Type Questions  1  20  20 
Short Answer Type Questions  I  2  6  12 
Short Answer Type Questions  II  3  8  24 
Long Answer Type Questions  4  6  24 
Total  40  80 
For Preparation of exams students can also check out other resource material
CBSE Class 9 Maths Sample Papers
CBSE Class 9 Maths Worksheets
CBSE Class 9 Maths Question Papers
CBSE Class 9 Maths Test Papers
CBSE Class 9 Maths Revision Notes
Question Bank of Other Subjects of Class 9
Importance of Question Bank for Exam Preparation?
There are many ways to ascertain whether a student has understood the important points and topics of a particular chapter and is he or she well prepared for exams and tests of that particular chapter. Apart from reference books and notes, Question Banks are very effective study materials for exam preparation. When a student tries to attempt and solve all the important questions of any particular subject , it becomes very easy to gauge how much well the topics have been understood and what kind of questions are asked in exams related to that chapter.. Some of the other advantaging factors of Question Banks are as follows
 Since Important questions included in question bank are collections of questions that were asked in previous exams and tests thus when a student tries to attempt them they get a complete idea about what type of questions are usually asked and whether they have learned the topics well enough. This gives them an edge to prepare well for the exam.Students get the clear idea whether the questions framed from any particular chapter are mostly either short or long answer type questions or multiple choice based and also marks weightage of any particular chapter in final exams.
 CBSE Question Banks are great tools to help in analysis for Exams. As it has a collection of important questions that were asked previously in exams thereby it covers every question from most of the important topics. Thus solving questions from the question bank helps students in analysing their preparation levels for the exam. However the practice should be done in a way that first the set of questions on any particular chapter are solved and then solutions should be consulted to get an analysis of their strong and weak points. This ensures that they are more clear about what to answer and what can be avoided on the day of the exam.
 Solving a lot of different types of important questions gives students a clear idea of what are the main important topics of any particular chapter that needs to focussed on from examination perspective and should be emphasised on for revision before attempting the final paper. So attempting most frequently asked questions and important questions helps students to prepare well for almost everything in that subject.
 Although students cover up all the chapters included in the course syllabus by the end of the session, sometimes revision becomes a time consuming and difficult process. Thus, practicing important questions from Question Bank allows students to check the preparation status of each and every small topic in a chapter. Doing that ensures quick and easy insight into all the important questions and topics in each and every individual. Solving the important questions also acts as the revision process.
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CBSE Case Study Questions for Class 9 Maths  Pdf PDF Download
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CBSE Case Study Questions for Class 9 Maths
CBSE Case Study Questions for Class 9 Maths are a type of assessment where students are given a realworld scenario or situation and they need to apply mathematical concepts to solve the problem. These types of questions help students to develop their problemsolving skills and apply their knowledge of mathematics to reallife situations.
Chapter Wise Case Based Questions for Class 9 Maths
The CBSE Class 9 Case Based Questions can be accessed from Chapetrwise Links provided below:
Chapterwise casebased questions for Class 9 Maths are a set of questions based on specific chapters or topics covered in the maths textbook. These questions are designed to help students apply their understanding of mathematical concepts to realworld situations and events.
Chapter 1: Number System
 Case Based Questions: Number System
Chapter 2: Polynomial
 Case Based Questions: Polynomial
Chapter 3: Coordinate Geometry
 Case Based Questions: Coordinate Geometry
Chapter 4: Linear Equations
 Case Based Questions: Linear Equations  1
 Case Based Questions: Linear Equations 2
Chapter 5: Introduction to Euclid’s Geometry
 Case Based Questions: Lines and Angles
Chapter 7: Triangles
 Case Based Questions: Triangles
Chapter 8: Quadrilaterals
 Case Based Questions: Quadrilaterals  1
 Case Based Questions: Quadrilaterals  2
Chapter 9: Areas of Parallelograms
 Case Based Questions: Circles
Chapter 11: Constructions
 Case Based Questions: Constructions
Chapter 12: Heron’s Formula
 Case Based Questions: Heron’s Formula
Chapter 13: Surface Areas and Volumes
 Case Based Questions: Surface Areas and Volumes
Chapter 14: Statistics
 Case Based Questions: Statistics
Chapter 15: Probability
 Case Based Questions: Probability
Weightage of Case Based Questions in Class 9 Maths
Why are Case Study Questions important in Maths Class 9?
 Enhance critical thinking: Case study questions require students to analyze a reallife scenario and think critically to identify the problem and come up with possible solutions. This enhances their critical thinking and problemsolving skills.
 Apply theoretical concepts: Case study questions allow students to apply theoretical concepts that they have learned in the classroom to reallife situations. This helps them to understand the practical application of the concepts and reinforces their learning.
 Develop decisionmaking skills: Case study questions challenge students to make decisions based on the information provided in the scenario. This helps them to develop their decisionmaking skills and learn how to make informed decisions.
 Improve communication skills: Case study questions often require students to present their findings and recommendations in written or oral form. This helps them to improve their communication skills and learn how to present their ideas effectively.
 Enhance teamwork skills: Case study questions can also be done in groups, which helps students to develop teamwork skills and learn how to work collaboratively to solve problems.
In summary, case study questions are important in Class 9 because they enhance critical thinking, apply theoretical concepts, develop decisionmaking skills, improve communication skills, and enhance teamwork skills. They provide a practical and engaging way for students to learn and apply their knowledge and skills to reallife situations.
Class 9 Maths Curriculum at Glance
The Class 9 Maths curriculum in India covers a wide range of topics and concepts. Here is a brief overview of the Maths curriculum at a glance:
 Number Systems: Students learn about the real number system, irrational numbers, rational numbers, decimal representation of rational numbers, and their properties.
 Algebra: The Algebra section includes topics such as polynomials, linear equations in two variables, quadratic equations, and their solutions.
 Coordinate Geometry: Students learn about the coordinate plane, distance formula, section formula, and slope of a line.
 Geometry: This section includes topics such as Euclid’s geometry, lines and angles, triangles, and circles.
 Trigonometry: Students learn about trigonometric ratios, trigonometric identities, and their applications.
 Mensuration: This section includes topics such as area, volume, surface area, and their applications.
 Statistics and Probability: Students learn about measures of central tendency, graphical representation of data, and probability.
The Class 9 Maths curriculum is designed to provide a strong foundation in mathematics and prepare students for higher education in the field. The curriculum is structured to develop critical thinking, problemsolving, and analytical skills, and to promote the application of mathematical concepts in reallife situations. The curriculum is also designed to help students prepare for competitive exams and develop a strong mathematical base for future academic and professional pursuits.
Students can also access Case Based Questions of all subjects of CBSE Class 9
 Case Based Questions for Class 9 Science
 Case Based Questions for Class 9 Social Science
 Case Based Questions for Class 9 English
 Case Based Questions for Class 9 Hindi
 Case Based Questions for Class 9 Sanskrit
Frequently Asked Questions (FAQs) on Case Based Questions for Class 9 Maths
What is casebased questions.
CaseBased Questions (CBQs) are openended problem solving tasks that require students to draw upon their knowledge of Maths concepts and processes to solve a novel problem. CBQs are often used as formative or summative assessments, as they can provide insights into how students reason through and apply mathematical principles in realworld problems.
What are casebased questions in Maths?
Casebased questions in Maths are problemsolving tasks that require students to apply their mathematical knowledge and skills to realworld situations or scenarios.
What are some common types of casebased questions in class 9 Maths?
Common types of casebased questions in class 9 Maths include word problems, realworld scenarios, and mathematical modeling tasks.
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 CBSE Maths Important Questions
 Class 9 Maths
 Chapter 2: Polynomial
Important Questions CBSE Class 9 Maths Chapter 2 Polynomial
Important questions for Class 9 Maths Chapter 2 Polynomials are provided here to help the CBSE students score well in their Class 9 Maths exam. The practice questions given here from polynomials chapter (NCERT) will help the students to create a better understanding of the concepts and, thus, develop their problemsolving skills.
Students can find the CBSE Class 9 Important questions from Chapter 2 Polynomials of the subject Maths here. These questions help students to be familiar with the question types and thus face the exam more confidently.
Also Check:
 Important 2 Marks Questions for CBSE 9th Maths
 Important 3 Marks Questions for CBSE 9th Maths
 Important 4 Marks Questions for CBSE 9th Maths
Important Polynomials Questions For Class 9 Chapter 2 (With Solutions)
Some important questions from polynomials are given below with solutions. These questions will help the 9th class students to get acquainted with a wide variety of questions and develop the confidence to solve polynomial questions more efficiently.
1. Give an example of a monomial and a binomial having degrees of 82 and 99, respectively.
An example of a monomial having a degree of 82 = x 82
An example of a binomial having a degree of 99 = x 99 + x
2. Compute the value of 9x 2 + 4y 2 if xy = 6 and 3x + 2y = 12.
Consider the equation 3x + 2y = 12
Now, square both sides:
(3x + 2y) 2 = 12 2
=> 9x 2 + 12xy + 4y 2 = 144
=>9x 2 + 4y 2 = 144 – 12xy
From the questions, xy = 6
9x 2 + 4y 2 = 144 – 72
Thus, the value of 9x 2 + 4y 2 = 72
3. Find the value of the polynomial 5x – 4x 2 + 3 at x = 2 and x = –1.
Let the polynomial be f(x) = 5x – 4x 2 + 3
Now, for x = 2,
f(2) = 5(2) – 4(2) 2 + 3
=> f(2) = 10 – 16 + 3 = –3
Or, the value of the polynomial 5x – 4x 2 + 3 at x = 2 is 3.
Similarly, for x = –1,
f(–1) = 5(–1) – 4(–1) 2 + 3
=> f(–1) = –5 –4 + 3 = 6
The value of the polynomial 5x – 4x 2 + 3 at x = 1 is 6.
4. Calculate the perimeter of a rectangle whose area is 25x 2 – 35x + 12.
Area of rectangle = 25x 2 – 35x + 12
We know, area of rectangle = length × breadth
So, by factoring 25x 2 – 35x + 12, the length and breadth can be obtained.
25x 2 – 35x + 12 = 25x 2 – 15x – 20x + 12
=> 25x 2 – 35x + 12 = 5x(5x – 3) – 4(5x – 3)
=> 25x 2 – 35x + 12 = (5x – 3)(5x – 4)
So, the length and breadth are (5x – 3)(5x – 4).
Now, perimeter = 2(length + breadth)
= 2(5x – 3 + 5x – 4) = 2(10x – 7) = 20x – 14
So, the perimeter = 20x – 14
5. Find the value of x 3 + y 3 + z 3 – 3xyz if x 2 + y 2 + z 2 = 83 and x + y + z = 15
Consider the equation x + y + z = 15
From algebraic identities, we know that (a + b + c) 2 = a 2 + b 2 + c 2 + 2(ab + bc + ca)
(x + y + z) 2 = x 2 + y 2 + z 2 + 2(xy + yz + xz)
From the question, x 2 + y 2 + z 2 = 83 and x + y + z = 15
15 2 = 83 + 2(xy + yz + xz)
=> 225 – 83 = 2(xy + yz + xz)
Or, xy + yz + xz = 142/2 = 71
Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca),
x 3 + y 3 + z 3 – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz))
x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71
So, x 3 + y 3 + z 3 – 3xyz = 15(83 – 71)
=> x 3 + y 3 + z 3 – 3xyz = 15 × 12
Or, x 3 + y 3 + z 3 – 3xyz = 180
6. If a + b + c = 15 and a 2 + b 2 + c 2 = 83, find the value of a 3 + b 3 + c 3 – 3abc.
We know that,
a 3 + b 3 + c 3 – 3abc = (a + b + c)(a 2 + b 2 + c 2 – ab – bc – ca) ….(i)
(a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca ….(ii)
Given, a + b + c = 15 and a 2 + b 2 + c 2 = 83
From (ii), we have
15 2 = 83 + 2(ab + bc + ca)
⇒ 225 – 83 = 2(ab + bc + ca)
⇒ 142/2 = ab + bc + ca
⇒ ab + bc + ca = 71
Now, (i) can be written as
a 3 + b 3 + c 3 – 3abc = (a + b + c)[(a 2 + b 2 + c 2 ) – (ab + bc + ca)]
a 3 + b 3 + c 3 – 3abc = 15 × [83 – 71] = 15 × 12 = 180.
7. If (x – 1/x) = 4, then evaluate (x 2 + 1/x 2 ) and (x 4 + 1/x 4 ).
Given, (x – 1/x) = 4
Squaring both sides we get,
(x – 1/x) 2 = 16
⇒ x 2 – 2.x.1/x + 1/x 2 = 16
⇒ x 2 – 2 + 1/x 2 = 16
⇒ x 2 + 1/x 2 = 16 + 2 = 18
∴ (x 2 + 1/x 2 ) = 18 ….(i)
Again, squaring both sides of (i), we get
(x 2 + 1/x 2 ) 2 = 324
⇒ x 4 + 2.x 2 .1/x 2 + 1/x 4 = 324
⇒ x 4 + 2 + 1/x 4 = 324
⇒ x 4 + 1/x 4 = 324 – 2 = 322
∴ (x 4 + 1/x 4 ) = 322.
8. Find the values of a and b so that (2x 3 + ax 2 + x + b) has (x + 2) and (2x – 1) as factors.
Let p(x) = 2x 3 + ax 2 + x + b. Then, p( –2) = and p(½) = 0.
p(2) = 2(2) 3 + a(2) 2 + 2 + b = 0
⇒ –16 + 4a – 2 + b = 0 ⇒ 4a + b = 18 ….(i)
p(½) = 2(½) 3 + a(½) 2 + (½) + b = 0
⇒ a + 4b = –3 ….(ii)
On solving (i) and (ii), we get a = 5 and b = –2.
Hence, a = 5 and b = –2.
9. Check whether (7 + 3x) is a factor of (3x 3 + 7x).
Let p(x) = 3x 3 + 7x and g(x) = 7 + 3x. Now g(x) = 0 ⇒ x = –7/3.
By the remainder theorem , we know that when p(x) is divided by g(x) then the remainder is p(–7/3).
Now, p(–7/3) = 3(–7/3) 3 + 7(–7/3) = –490/9 ≠ 0.
∴ g(x) is not a factor of p(x).
10. Factorise x 2 + 1/x 2 + 2 – 2x – 2/x.
Solution:
x 2 + 1/x 2 + 2 – 2x – 2/x = (x 2 + 1/x 2 + 2) – 2(x + 1/x)
= (x + 1/x) 2 – 2(x + 1/x)
= (x + 1/x)(x + 1/x – 2).
11. Factorise x 2 – 1 – 2a – a 2 .
x 2 – 1 – 2a – a 2 = x 2 – (1 + 2a + a 2 )
= x 2 – (1 + a) 2
= [x – (1 – a)][x + 1 + a]
= (x – 1 – a)(x + 1 + a)
∴ x 2 – 1 – 2a – a 2 = (x – 1 – a)(x + 1 + a).
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its very nice app
tough please make the solution a bit easy
I thank byjus for providing these important questions
Nice questions but little bit difficult make it easy to solve
I don’t got question 5, step x^3+y^3+z^33xyz=15(8371) Why we subtracted 8371??
We evaluated the value of xy + yz + xz, initially, which is equal to 71. Since, x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz)) [By algebraic identities] And x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71 So, if we substitute the values, we get: x^3 + y^3 + z^3 – 3xyz = 15(83 – 71) Please go through the complete solution for the answer.
The expression is x^3+y^3+z^3 = (x+y+z)(x^2+y^2+z^2xyyzxz)
= (x+y+z){(x^2+y^2+z^2)(xy+yz+sa)}
x^2 +y^2 + z^2 = 83, xy+yz+sa = 71 This is why 83 71 is done
x + y + z = 15, x² + y² + z² = 83 And xy + yz + xz = 71 Now, Acc. To the question, x3 + y3 + z3 – 3xyz which is equal to = (x + y + z)(x² + y² + z² – (xy + yz + xz)) And these are given above So we subtract 83 – 71
because the formula is x^3+y^3+z^33xyz = (x+y+z)(x^2 + y^2 + z^2xyyzzx) and we know x+y+z= 15 and x^2 + y^2 + z^2 = 83 and xy+yz+zx = 71 and xyyzzx = 71 so we have putted the values on formula 15(8371)
we found the value of xy+yz+zx so in the above step we take ‘‘ as common and then xy+yz+zx so it will 71 then it is 15*12=180
This is so nice we can learn easily
wow! nice app
Thank you so much for these questions
Nice questions and its also easy to check since the solutions are given
Very helpful during online education
maths in byjus is easy. thank you for everything
Please add some more questions. There should be at least 10 questions
very good question for the exam. thank you
Nice app but difficult to do the question plz make it
Very easy to solve
Very nice app and there question is very useful to me, there question is very interesting .
These are the most basic and the most common questions for exams. Thank you Byjus app for making studies easier than ever.
These are the most basic , the most common and the most tough questions for exams. Thank you Byjus app for making studies easier than ever.
thank you for providing these important questions
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 CBSE Class 9 Practice Papers
CBSE Class 9 Maths: Extra Questions  Chapter 2  Polynomials (with Answers)
Check here the ncert based extra question for cbse class 9 maths chapter 2 polynomials. all these questions are important for the annual cbse exam..
CBSE Class 9 Maths extra questions (with answers) for Chapter 2  Polynomials can be accessed from here. These extra questions are entirely based on the NCERT textbook. Most of the questions are simple and appropriate to test your conceptual understanding. So, you should solve all these questions for selfassessment and score well in your Maths Exam 20202021.
CBSE Class 9 Maths Extra Questions for Chapter 2  Polynomials :
1. Degree of a zero polynomial is
(iii) Not defined
Not defined
2. What is the degree of a constant polynomial?
3. Number of zeroe(s) all the linear polynomial have is/are:
(iii) Three
4. Which of the following expressions is not a polynomial?
(i) 5x 3 +x 2 –2
(ii) x 1/2 +3x+1
(iii) 3x 1 +1
(iv) (9x +1) ÷ (x)
(i) 5x 3 +x 2 2
5. Find the product of (x – 3y) (x + 3y) (x 2 + 9y 2 ).
(x 4 – 81y 4 )
Also Check:
NCERT Book for Class 9 Maths
NCERT Solutions for Class 9 Maths
6. Find the value of x – 1/x, If x 2 + 1/x 2 = 18.
x – 1/x = ±4
7. Find the value of 105 × 106 without actual multiplication?
Answer:
105 × 106 = 11,130
8. Which of the following statements (s) is/are correct?
(i) Every linear polynomial in one variable has a unique zero.
(ii) Every nonzero constant polynomial has no zero.
(iii) Every real number is a zero of the zero polynomial.
(iv) All of the above statements are correct
9. Factorise: (a – b) 3 + (b – c) 3 + (c – a) 3
3(a 2 (c−b) + b 2 (a−c) + c 2 (b−a))
10. If p(x) is a polynomial of degree n > 1 and a is any real number, then (i) x – a is a factor of p(x),
(i) If p(a) > 0
(ii) If p(a) < 0
(iii) If p(a) = 0
(iv) For any value of P(a)
If p(a) = 0
Students must go through the latest CBSE Syllabus for Class 9 Maths so that they can prepare according to the contents prescribed by the board.
Also check:
CBSE Class 9 Maths Important Questions and Answers
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 NCERT Exemplar for Class 9 Maths Chapter 2  Polynomials (Book Solutions)
 Textbook Solutions
NCERT Exemplar for Class 9 Maths  Polynomials  Free PDF Download
Free PDF download of NCERT Exemplar for Class 9 Maths Chapter 2  Polynomials solved by expert Maths teachers on Vedantu as per NCERT (CBSE) Book guidelines. All Chapter 2  Polynomials exercise questions with solutions to help you to revise the complete syllabus and score more marks in your examinations. Download free Class 9 Maths to amp up your preparations and to score well in your examinations. Students can also avail of NCERT Solutions Class 9 Science from our website. Besides, find NCERT Solutions to get more understanding of various subjects. The solutions are uptodate and are sure to help in your academic journey.
Access NCERT Exemplar Solutions for Class 9 Mathematics Chapter 2  Polynomials
Multiple choice questions.
Sample Question 1: If \[{x^2}{\text{ }} + {\text{ }}kx{\text{ }} + {\text{ }}6{\text{ }} = {\text{ }}\left( {x{\text{ }} + {\text{ }}2} \right){\text{ }}\left( {x{\text{ }} + {\text{ }}3} \right)\] for all x, then the value of k is
(A) $ 1 $
(B) $  1 $
(C) $ 5 $
(D) $ 3 $
Ans: Option (C) is correct.
We have, \[{x^2}{\text{ }} + {\text{ }}kx{\text{ }} + {\text{ }}6{\text{ }} = {\text{ }}\left( {x{\text{ }} + {\text{ }}2} \right){\text{ }}\left( {x{\text{ }} + {\text{ }}3} \right)\]
Now, let’s taking RHS,
$ \Rightarrow \left( {x + 2} \right)\left( {x + 3} \right) $
On multiplication, we get
$ \Rightarrow {x^2} + 2x + 3x + 6 $
On addition, we get
$ \Rightarrow {x^2} + 5x + 6 $
Now on comparing LHS and RHS, we get
$ k = 5 $
EXERCISE 2.1
Write the correct option in each of the following:
1. Which of the following is a polynomial?
(A) $ \dfrac{{{x^2}}}{2}  \dfrac{2}{{{x^2}}} $
(B) $ \sqrt {2x}  1 $
(C) $ {x^2} + \dfrac{{3{x^{\dfrac{3}{2}}}}}{{\sqrt x }} $
(D) $ \dfrac{{x  1}}{{x + 1}} $
Ans: (C) is the correct option.
Polynomials are an algebraic expression that consist of variables and coefficients. Here the power of variables should be in whole number like 1,2,3 etc.
(A) $ \dfrac{{{x^2}}}{2}  \dfrac{2}{{{x^2}}} $
Here the coefficient $  2 $ is multiplied with $ {x^{  2}} $ . So here the power of x is not a whole number. Hence, it is not a Polynomial.
Here the coefficient $ \sqrt 2 $ is multiplied with $ {x^{\dfrac{1}{2}}} $ . So here the power of x is not a whole number. Hence, it is not a Polynomial.
Here the coefficient $ 1 $ is multiplied with $ {x^2} $ and the coefficient $ 3 $ is multiplied with \[{x^{\dfrac{3}{2}  \dfrac{1}{2}}} = {x^{\dfrac{2}{2}}} = x\] . So here the power of x is a whole number in the whole equation. Hence, it is a Polynomial.
It is not a standard format of a Polynomial. Hence, it is not a Polynomial.
2. $ \sqrt 2 $ is a Polynomial of degree
(A) $ 2 $
(B) $ 0 $
(C) $ 1 $
(D) $ \dfrac{1}{2} $
Ans: Option (B) is correct.
Here $ \sqrt 2 $ is a constant Polynomial. We can also write it as $ \sqrt 2 {x^0} $ . Therefore, the degree of this Polynomial is $ 0 $ .
3. Degree of the Polynomial of $ 4{x^4} + 0{x^3} + 0{x^5} + 5x + 7 $ is
(A) $ 4 $
(B) $ 5 $
(C) $ 3 $
(D) $ 7 $
Ans: Option (A) is correct.
The highest power of x in the Polynomial is called the degree of Polynomial. We have the term with the highest power of x is $ 4{x^4} $ , which is $ 4 $ .
Therefore, the degree of Polynomial is $ 4 $ .
4. Degree of the zero Polynomial
(A) $ 0 $
(B) $ 1 $
(C) Any natural number
(D) Not defined
Ans: Option (D) is the correct answer.
Like any constant value, the value 0 can be considered as a (constant) Polynomial, called the zero Polynomial. It has no nonzero terms, and so, strictly speaking, it has no degree either. Therefore, the degree of the zero Polynomial is not defined.
5. If $ p\left( x \right) = {x^2}  2\sqrt 2 x + 1 $ , then $ p\left( {2\sqrt 2 } \right) $ is equal to
(A) $ 0 $
(C) $ 4\sqrt 2 $
(D) $ 8\sqrt 2 + 1 $
Ans: The correct option is (B).
$ p\left( x \right) = {x^2}  2\sqrt 2 x + 1 $
Here, we have given the value of $ x = 2\sqrt 2 $ .
So, on putting the value in above equation, we get
$ p\left( {2\sqrt 2 } \right) = {\left( {2\sqrt 2 } \right)^2}  2\sqrt 2 \left( {2\sqrt 2 } \right) + 1 $
$ p\left( {2\sqrt 2 } \right) = 4 \times 2  4 \times 2 + 1 $
$ p\left( {2\sqrt 2 } \right) = 8  8 + 1 $
$ p\left( {2\sqrt 2 } \right) = 1 $
6. The value of the Polynomial $ 5x  4{x^2} + 3 $ , when $ x =  1 $ is
(A) $  6 $
(B) $ 6 $
(C) $ 2 $
(D) $  2 $
Ans: The correct option is (A).
Here we have given a Polynomial $ 5x  4{x^2} + 3 $ .
In this we have to put $ x =  1 $ , and then we have to find the value of Polynomial
$ p\left( x \right) = 5x  4{x^2} + 3 $
$ p\left( {  1} \right) = 5\left( {  1} \right)  4{\left( {  1} \right)^2} + 3 $
$ p\left( {  1} \right) =  5  4 + 3 $
$ p\left( {  1} \right) =  9 + 3 $
$ p\left( {  1} \right) =  6 $
7. If $ p\left( x \right) = x + 3 $ , then $ p\left( x \right) + p\left( {  x} \right) $ is equal to
(A) $ 3 $
(B) $ 2x $
(C) $ 0 $
(D) $ 6 $
Ans: The correct option is (D).
Here, we have $ p\left( x \right) = x + 3 $ .
Now, on substituting $ x $ with $  x $ , we get $ p\left( {  x} \right) =  x + 3 $
Now, on adding both the equations, we get
$ p\left( x \right) + p\left( {  x} \right) = x + 3  x + 3 $
$ p\left( x \right) + p\left( {  x} \right) = 6 $
8. Zero of a zero Polynomial is
(C) Any real number
Ans: (C) is the correct answer.
Zero Polynomial is a constant Polynomial whose all coefficients are equal to 0. Zero of a Polynomial or we can say that root of a polynomial is the value of a variable that is responsible to make the Polynomial equals to zero. Hence, zero of the zero Polynomial is any real number.
9. Zero of the Polynomial $ p\left( x \right) = 2x + 5 $ is
(A) $  \dfrac{2}{5} $
(B) $  \dfrac{5}{2} $
(C) $ \dfrac{2}{5} $
(D) $ \dfrac{5}{2} $
Ans : (B) is the correct option.
To get the zero of the Polynomial, we have to put the expression equals to zero.
$ p\left( x \right) = 0 $
$ 2x + 5 = 0 $
$ 2x =  5 $
$ x =  \dfrac{5}{2} $
Therefore, $  \dfrac{5}{2} $ is the zero of the Polynomial.
10. One of the zeroes of given Polynomial $ 2{x^2} + 7x  4 $ is
(A) $ 2 $
(B) $ \dfrac{1}{2} $
(C) $  \dfrac{1}{2} $
(D) $  2 $
Ans: (B) is the correct option.
To find the zeroes of a Polynomial, we have to put that expression equals to zero.
$ 2{x^2} + 7x  4 = 0 $
Using middle term splitting,
$ 2{x^2} + \left( {8  1} \right)x  4 = 0 $
$ 2{x^2} + 8x  x  4 = 0 $
$ 2x\left( {x + 4} \right)  1\left( {x + 4} \right) = 0 $
$ \left( {2x  1} \right)\left( {x + 4} \right) = 0 $
So, there we have two Ansutions
$ 2x  1 = 0 $ and $ x + 4 = 0 $
$ 2x = 1 $ and $ x =  4 $
$ x = \dfrac{1}{2} $ and $ x =  4 $
11. If $ {x^{51}} + 51 $ is divided by $ x + 1 $ , the remainder is
(A) $ 0 $
(B) $ 1 $
(C) $ 49 $
(D) $ 50 $
Ans: Option (D) is the correct option.
Here we will use the remainder theorem.
If any Polynomial $ f(x) $ is divided by $ x  h $ , then the remainder will be $ f\left( h \right) $
Here given Polynomial is $ {x^{51}} + 51 $ and divided by $ x + 1 $
So, by using the remainder theorem we can say that
Remainder= $ f\left( {  1} \right) $ since the divisor is $ x + 1 $
Therefore we can write it as
\[ \Rightarrow f(x) = {x^{51}} + 51\]
\[ \Rightarrow f(  1) = {(  1)^{51}} + 51\]
\[ \Rightarrow f(  1) =  1 + 51\]
\[\therefore f(  1) = 50\]
12. If $ x + 1 $ , is a factor of the Polynomial $ 2{x^2} + kx $ , then the value of k is
(A) $  3 $
(B) $ 4 $
(C) $ 2 $
Ans: Option (C) is correct.
Let $ p\left( x \right) = 2{x^2} + kx $
Since, \[\left( {x{\text{ }} + {\text{ }}1} \right)\] is a factor of \[p\left( x \right)\] , then
Using, factor theorem
\[ \Rightarrow p\left( {  1} \right) = 0\]
\[ \Rightarrow 2{\left( {  1} \right)^2} + {\text{ }}k\left( {  1} \right){\text{ }} = {\text{ }}0\]
\[ \Rightarrow \;2  k{\text{ }} = {\text{ }}0\]
\[ \Rightarrow \;k = {\text{ }}2\]
Therefore, the value of k is $ 2 $ .
13. $ x + 1 $ , is a factor of the Polynomial
(A) $ {x^3} + {x^2}  x + 1 $
(B) $ {x^3} + {x^2} + x + 1 $
(C) $ {x^4} + {x^3} + {x^2} + 1 $
(D) $ {x^4} + 3{x^3} + 3{x^2} + x + 1 $
If $ x + 1 $ is the factor of the Polynomial, then $ x =  1 $ is the root of the Polynomial.
(A) Let $ p\left( x \right) = {x^3} + {x^2}  x + 1 $
Now, put $ x =  1 $ .
$ p\left( {  1} \right) = {\left( {  1} \right)^3} + {\left( {  1} \right)^2}  \left( {  1} \right) + 1 $
$ p\left( {  1} \right) =  1 + 1 + 1 + 1 $
$ p\left( {  1} \right) = 2 $
Therefore, $ p\left( {  1} \right) \ne 0 $ .
So, $ x + 1 $ is not a factor of the $ p\left( x \right) $ .
(B) Let $ p\left( x \right) = {x^3} + {x^2} + x + 1 $
$ p\left( {  1} \right) = {\left( {  1} \right)^3} + {\left( {  1} \right)^2} + \left( {  1} \right) + 1 $
$ p\left( {  1} \right) =  1 + 1  1 + 1 $
$ p\left( {  1} \right) = 0 $
Therefore, $ p\left( {  1} \right) = 0 $ .
So, $ x + 1 $ is a factor of the $ p\left( x \right) $ .
(C) Let $ p\left( x \right) = {x^4} + {x^3} + {x^2} + 1 $
$ p\left( {  1} \right) = {\left( {  1} \right)^4} + {\left( {  1} \right)^3} + {\left( {  1} \right)^2} + 1 $
$ p\left( {  1} \right) = 1  1 + 1 + 1 $
(D) Let $ p\left( x \right) = {x^4} + 3{x^3} + 3{x^2} + x + 1 $
$ p\left( {  1} \right) = {\left( {  1} \right)^4} + 3{\left( {  1} \right)^3} + 3{\left( {  1} \right)^2} + \left( {  1} \right) + 1 $
$ p\left( {  1} \right) = 1  3 + 3  1 + 1 $
$ p\left( {  1} \right) = 1 $
So, $ x + 1 $ is not the factor of the $ p\left( x \right) $ .
14. One of the factor of $ \left( {25{x^2}  1} \right) + {\left( {1 + 5x} \right)^2} $ is
(A) $ 5 + x $
(B) $ 5  x $
(C) $ 5x  1 $
(D) $ 10x $
Ans: (D) is the correct option.
$ \Rightarrow \left( {25{x^2}  1} \right) + {\left( {1 + 5x} \right)^2} $
$ \Rightarrow \left( {{{\left( {5x} \right)}^2}  {{\left( 1 \right)}^2}} \right) + {\left( {5x + 1} \right)^2} $
$ \Rightarrow \left( {5x + 1} \right)\left( {5x  1} \right) + {\left( {5x + 1} \right)^2} $
Now, taking common $ \left( {5x + 1} \right) $
$ \Rightarrow \left( {5x + 1} \right)\left( {\left( {5x  1} \right) + \left( {5x + 1} \right)} \right) $
$ \Rightarrow \left( {5x + 1} \right)\left( {5x  1 + 5x + 1} \right) $
$ \Rightarrow \left( {5x + 1} \right)\left( {10x} \right) $
Therefore, one of the factor of $ \left( {25{x^2}  1} \right) + {\left( {1 + 5x} \right)^2} $ is $ 10x $ .
15. The value of $ {249^2}  {248^2} $ is
(A) $ {1^2} $
(B) $ 477 $
(C) $ 487 $
(D) $ 497 $
Here we will use the identity $ {a^2}  {b^2} = \left( {a + b} \right)\left( {a  b} \right) $ .
We have,
$ \Rightarrow {249^2}  {248^2} $
$ \Rightarrow \left( {249  248} \right)\left( {249 + 248} \right) $
$ \Rightarrow \left( 1 \right)\left( {497} \right) $
$ \Rightarrow 497 $
16. The factorization of $ 4{x^2} + 8x + 3 $ is
(A) $ \left( {x + 1} \right)\left( {x + 3} \right) $
(B) $ \left( {2x + 1} \right)\left( {2x + 3} \right) $
(C) $ \left( {2x + 2} \right)\left( {2x + 5} \right) $
(D) $ \left( {2x  1} \right)\left( {2x  3} \right) $
In this question, we will use middle term splitting for factorization
$ \Rightarrow 4{x^2} + 8x + 3 $
We can also write it as,
$ \Rightarrow 4{x^2} + 6x + 2x + 3 $
$ \Rightarrow 2x\left( {2x + 3} \right) + 1\left( {2x + 3} \right) $
$ \Rightarrow \left( {2x + 1} \right)\left( {2x + 3} \right) $
17. Which of the followings is factor of $ {\left( {x + y} \right)^3}  \left( {{x^3} + {y^3}} \right) $ ?
(A) $ {x^2} + {y^2} + 2xy $
(B) $ {x^2} + {y^2}  xy $
(C) $ x{y^2} $
(D) $ 3xy $
We have, $ {\left( {x + y} \right)^3}  \left( {{x^3} + {y^3}} \right) $ .
Now, in this question we will use the identity $ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right) $
$ \Rightarrow {x^3} + {y^3} + 3xy\left( {x + y} \right)  \left( {{x^3} + {y^3}} \right) $
$ \Rightarrow 3xy\left( {x + y} \right) $
So, we can say that $ 3xy $ is a factor of $ {\left( {x + y} \right)^3}  \left( {{x^3} + {y^3}} \right) $ .
18. The coefficient of variable x in the expansion $ {\left( {x + 3} \right)^3} $ is
(B) $ 9 $
(C) $ 18 $
(D) $ 27 $
Here in this question, we will use the identity $ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right) $
Therefore,
$ {\left( {x + 3} \right)^3} = {x^3} + {3^3} + 3 \times x \times 3\left( {x + 3} \right) $
$ {\left( {x + 3} \right)^3} = {x^3} + 27 + 9{x^2} + 27x $
Therefore, the coefficient of x is $ 27 $ .
19. If $ \dfrac{x}{y} + \dfrac{y}{x} =  1 $ the value of $ {x^3}  {y^3} $ is
Ans: The correct option is (C).
Here, it is given that $ \dfrac{x}{y} + \dfrac{y}{x} =  1 $
On taking LCM, we get
$ \dfrac{{{x^2} + {y^2}}}{{xy}} =  1 $
On crossmultiplication, we get
$ {x^2} + {y^2} =  xy $
Now, we know that
$ {x^3}  {y^3} = \left( {x  y} \right)\left( {{x^2} + {y^2} + xy} \right) $
On putting $ {x^2} + {y^2} =  xy $ in above equation, we get
$ {x^3}  {y^3} = \left( {x  y} \right)\left( {  xy + xy} \right) $
$ {x^3}  {y^3} = \left( {x  y} \right)\left( 0 \right) $
$ {x^3}  {y^3} = 0 $
20. If $ 49{x^2}  b = \left( {7x + \dfrac{1}{2}} \right)\left( {7x  \dfrac{1}{2}} \right) $ , then the value of b is
(A) $ 0 $
(B) $ \dfrac{1}{{\sqrt 2 }} $
(C) $ \dfrac{1}{4} $
Here in this question, we will use the identity $ {a^2}  {b^2} = \left( {a + b} \right)\left( {a  b} \right) $
$ \Rightarrow 49{x^2}  b = \left( {7x + \dfrac{1}{2}} \right)\left( {7x  \dfrac{1}{2}} \right) $
$ \Rightarrow 49{x^2}  b = {\left( {7x} \right)^2}  {\left( {\dfrac{1}{2}} \right)^2} $
$ \Rightarrow 49{x^2}  b = 49{x^2}  {\left( {\dfrac{1}{2}} \right)^2} $
On comparing, we get
$ b = {\left( {\dfrac{1}{2}} \right)^2} $
$ b = \dfrac{1}{4} $
21. If $ a + b + c = 0 $ , then the value of $ {a^3} + {b^3} + {c^3} $ is equal to
(B) $ abc $
(C) $ 3abc $
(D) $ 2abc $
Here we know an identity,
$ {a^3} + {b^3} + {c^3}  3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2}  ab  bc  ca} \right) $
It is given that, $ a + b + c = 0 $
Therefore, put it in the above equation
$ \Rightarrow {a^3} + {b^3} + {c^3}  3abc = \left( 0 \right)\left( {{a^2} + {b^2} + {c^2}  ab  bc  ca} \right) $
$ \Rightarrow {a^3} + {b^3} + {c^3}  3abc = 0 $
$ \Rightarrow {a^3} + {b^3} + {c^3} = 3abc $
Short Answer Questions with Reasoning
Sample Question 1: Write whether the following statements are True or False.
Justify your answer .
(i) $ \dfrac{1}{{\sqrt 5 }}{x^{\dfrac{1}{2}}} + 1 $ is a polynomial
Ans: It is a false statement.
Here, the exponent of a variable is not a whole number. So, it is not a polynomial.
(ii) $ \dfrac{{6\sqrt x + {x^{\dfrac{3}{2}}}}}{{\sqrt x }} $ is a polynomial, $ x \ne 0 $
Ans: It is a true statement.
We can also write it as,
$ \dfrac{{6\sqrt x + {x^{\dfrac{3}{2}}}}}{{\sqrt x }} = \dfrac{{\sqrt x {{\left( {6 + x} \right)}^{}}}}{{\sqrt x }} = 6 + x $
Here, the exponent of variable is not a whole number. So, it is a polynomial.
EXERCISE 2.2
1. Which of the following expressions are Polynomials? Justify your answer.
(i) $ 8 $
Ans: Here we have given an expression $ 8 $ in which the degree of variable is zero. Therefore, it is a constant Polynomial.
(ii) $ \sqrt 3 {x^2}  2x $
Ans: $ \sqrt 3 {x^2}  2x $
Here, in each term the power of x is a whole number. Therefore, this expression is a Polynomial.
(iii) $ 1  \sqrt {5x} $
Ans: $ 1  \sqrt {5x} $
We can also write it as $ 1  \sqrt 5 {x^{\dfrac{1}{2}}} $ . Here the degree of variable is not a whole number. Therefore, this expression is not a Polynomial.
(iv) $ \dfrac{1}{{5{x^{  2}}}} + 5x + 7 $
Ans: $ \dfrac{1}{{5{x^{  2}}}} + 5x + 7 $
We can also write it as $ \dfrac{{1{x^2}}}{5} + 5x + 7 $ . Here in this expression each variable term has the power of x in a whole number. Therefore, this expression is a Polynomial.
(v) $ \dfrac{{\left( {x  2} \right)\left( {x  4} \right)}}{x} $
Ans: $ \dfrac{{\left( {x  2} \right)\left( {x  4} \right)}}{x} $
We can also write it as
$ \dfrac{{{x^2}  6x + 8}}{x} = x  6 + \dfrac{8}{x} = x  6 + 8{x^{  1}} $
Here, the exponent of variable x in the third term $ 8{x^{  1}} $ is – 1, and this is not a whole number.
Therefore, this algebraic expression is not a Polynomial.
(vi) $ \dfrac{1}{{x + 1}} $
Ans: $ \dfrac{1}{{x + 1}} $
We can also reduce it to $ {\left( {x + 1} \right)^{  1}} $ , which cannot be reduced to an expression in which the exponent of the variable x had only whole numbers in each of its terms. Therefore, this algebraic expression is not a Polynomial.
(vii) $ \dfrac{1}{7}{a^3}  \dfrac{2}{{\sqrt 3 }}{a^2} + 4a  7 $
Ans: $ \dfrac{1}{7}{a^3}  \dfrac{2}{{\sqrt 3 }}{a^2} + 4a  7 $
In this expression, the exponent of a in each term is a whole number, so this expression is a Polynomial.
(viii) $ \dfrac{1}{{2x}} $
Ans: $ \dfrac{1}{{2x}} $
We can also write it as $ \dfrac{1}{2}{x^{  1}} $ . Here, the exponent of the variable x is $  1 $ , which is not a whole number so, this algebraic expression is not a Polynomial.
2. Write whether the following statements are True or False. Justify your answer.
(i) A binomial can have at most two terms
Ans: The given statement is False because binomials have exactly two terms.
(ii) Every Polynomial is a binomial
Ans: Every Polynomial cannot be a binomial because a Polynomial had many terms but a binomial had only two terms in the expression. For example, $ {x^4} + {x^3} + {x^2} + x + 1 $ is a Polynomial but not a binomial. Therefore, the given statement is false.
(iii) A binomial may have degree $ 5 $
Ans: The given statement is True because a binomial is a Polynomial whose degree is a whole number and it is greater than or equals to one and contain two terms. For example, $ {x^5}  1 $ is a binomial of degree $ 5 $ .
(iv) Zero of a Polynomial is always $ 0 $
Ans: The given statement is False, because zero of Polynomial can be any real number.
(v) A Polynomial can't have more than one zeros
Ans: The given statement is False, because a Polynomial can have any number of zeroes which depends on the degree of the Polynomial.
(vi) The power of the sum of two polynomials each of whose degree $ 5 $ is always $ 5 $ .
Ans: The given statement is False. For example, consider the two Polynomials $  {x^5} + 3{x^2} + 4 $ and $ {x^5} + {x^4} + 2{x^3} + 3 $ .The degree of each of these polynomials will be 5. Their sum is $ {x^4} + 2{x^3} + 3{x^2} + 7 $ . The degree of this Polynomial is $ 4 $ not $ 5 $ .
Short Answer Questions
Sample Question 1: (i) Check whether p(x) is a multiple of g(x) or not, where \[\;p\left( x \right) = {x^3}{\text{ }}x + 1,\,{\text{ }}g\left( x \right) = 23x\]
Ans: p(x) will be a multiple of g(x) if and only if g(x) divides p(x).
Now, $ g\left( x \right) = 2  3x $
But, $ g\left( x \right) = 0 $
$ \Rightarrow 2  3x = 0 $
$ \Rightarrow x = \dfrac{2}{3} $
Now, put $ x = \dfrac{2}{3} $ in equation \[\;p\left( x \right) = {x^3}{\text{ }}x + 1\]
\[\;p\left( {\dfrac{2}{3}} \right) = {\left( {\dfrac{2}{3}} \right)^3}{\text{ }}\left( {\dfrac{2}{3}} \right) + 1\]
\[\;p\left( {\dfrac{2}{3}} \right) = \dfrac{8}{{27}} + \dfrac{1}{3} = \dfrac{{8 + 9}}{{27}} = \dfrac{{17}}{{27}}\]
Here, the remainder is not equals to zero.
Therefore, g(x) is not a multiple of p(x).
(ii) Check whether g(x) is a factor of p(x) or not, where \[p\left( x \right) = {\text{ }}8{x^3}6{x^2}4x + 3,{\text{ }}g\left( x \right) = {\text{ }}\dfrac{x}{3}  \dfrac{1}{4}\]
Now, $ g\left( x \right) = \dfrac{x}{3}  \dfrac{1}{4} $
$ \Rightarrow \dfrac{x}{3}  \dfrac{1}{4} = 0 $
$ \Rightarrow x = \dfrac{3}{4} $
Now, put $ x = \dfrac{3}{4} $ in equation \[p\left( x \right) = {\text{ }}8{x^3}6{x^2}4x + 3\]
\[p\left( {\dfrac{3}{4}} \right) = {\text{ }}8{\left( {\dfrac{3}{4}} \right)^3}6{\left( {\dfrac{3}{4}} \right)^2}4\left( {\dfrac{3}{4}} \right) + 3\]
\[p\left( {\dfrac{3}{4}} \right) = {\text{ }}8\left( {\dfrac{{27}}{{64}}} \right)6\left( {\dfrac{9}{{16}}} \right)3 + 3\]
\[p\left( {\dfrac{3}{4}} \right) = {\text{ }}\left( {\dfrac{{27}}{8}} \right)\left( {\dfrac{{27}}{8}} \right)3 + 3 = 0\]
Here, the remainder is equals to zero.
Therefore, g(x) is a multiple of p(x).
Sample Question 2: Find the value of a, if \[xa\] is a factor of \[{x^3}a{x^2}{\text{ }} + {\text{ }}2x{\text{ }} + {\text{ }}a{\text{ }}{\text{ }}1\].
Ans: Here, \[p\left( x \right) = {x^3}a{x^2}{\text{ }} + {\text{ }}2x{\text{ }} + {\text{ }}a{\text{ }}{\text{ }}1\]
If \[xa\] is a factor of p(x). Then, $ p\left( a \right) = 0 $
\[ \Rightarrow {a^3}a{\left( a \right)^2}{\text{ }} + {\text{ }}2\left( a \right){\text{ }} + {\text{ }}a{\text{ }}{\text{ }}1 = 0\]
$ \Rightarrow 3a  1 = 0 $
$ \Rightarrow a = \dfrac{1}{3} $
Therefore, $ a = \dfrac{1}{3} $ .
Sample Question 3: (i) Without actually calculating the cubes, find the value of \[{48^3}{\text{ }}{\text{ }}{30^3}{\text{ }}{\text{ }}{18^3}\].
Ans: We know that, \[{x^3}{\text{ }} + {\text{ }}{y^3}{\text{ }} + {\text{ }}{z^3}{\text{ }}{\text{ }}3xyz{\text{ }} = {\text{ }}\left( {x{\text{ }} + {\text{ }}y{\text{ }} + {\text{ }}z} \right){\text{ }}\left( {{x^2}{\text{ }} + {\text{ }}{y^2}{\text{ }} + {\text{ }}{z^2}{\text{ }}{\text{ }}xy{\text{ }}{\text{ }}yz{\text{ }}{\text{ }}zx} \right)\]
If, $ x + y + z = 0 $
Then, \[{x^3}{\text{ }} + {\text{ }}{y^3}{\text{ }} + {\text{ }}{z^3}{\text{ = }}3xyz{\text{ }}\]
Here, $ 48  30  18 = 0 $
Therefore, \[{48^3}{\text{ }}{\text{ }}{30^3}{\text{ }}{\text{ }}{18^3} = 3\left( {48} \right)\left( {  30} \right)\left( {  18} \right) = 77760\]
(ii)Without finding the cubes, factorise \[{\left( {x{\text{ }}{\text{ }}y} \right)^3}{\text{ }} + {\text{ }}{\left( {y{\text{ }}{\text{ }}z} \right)^3}{\text{ }} + {\text{ }}{\left( {z{\text{ }}{\text{ }}x} \right)^3}\] .
Here, $ \left( {x  y} \right) + \left( {y  z} \right) + \left( {z  x} \right) = 0 $
Therefore, \[{\left( {x{\text{ }}{\text{ }}y} \right)^3}{\text{ }} + {\text{ }}{\left( {y{\text{ }}{\text{ }}z} \right)^3}{\text{ }} + {\text{ }}{\left( {z{\text{ }}{\text{ }}x} \right)^3} = 3\left( {x  y} \right)\left( {y  z} \right)\left( {z  x} \right)\]
EXERCISE 2.3
1. Classify the following Polynomial as Polynomials in one variable, two variables etc.
(i) $ {x^2} + x + 1 $
Ans: $ {x^2} + x + 1 $ is a Polynomial in one variable.
(ii) $ {y^3}  5y $
Ans: $ {y^3}  5y $ is a Polynomial in one variable.
(iii) $ xy + yz + zx $
Ans: $ xy + yz + zx $ is a Polynomial in three variables.
(iv) $ {x^2}  2xy + {y^2} + 1 $
Ans: $ {x^2}  2xy + {y^2} + 1 $ is a Polynomial in two variables.
2. Determine the degree of each of the following Polynomials:
(i) $ 2x  1 $
Ans: Since the highest power of x is $ 1 $ , the degree of the Polynomial $ 2x  1 $ is $ 1 $ .
(ii) $  10 $
Ans: $  10 $ is a nonzero constant. A constant term does not contain any variable and its degree is always 0.
(iii) $ {x^3}  9x + 3{x^5} $
Ans: Since the highest power of x is $ 5 $ , the degree of the Polynomial $ {x^3}  9x + 3{x^5} $ is $ 5 $ .
(iv) $ {y^3}\left( {1  {y^4}} \right) $
Ans: Here, we have $ {y^3}\left( {1  {y^4}} \right) = {y^3}  {y^7} $ . Since the highest power of y is $ 7 $ , the degree of the Polynomial is $ 7 $ .
3. For the Polynomial $ \dfrac{{{x^3} + 2x + 1}}{5}  \dfrac{7}{2}{x^2}  {x^6} $, write
(i) the degree of the Polynomial
Ans: $ \dfrac{{{x^3} + 2x + 1}}{5}  \dfrac{7}{2}{x^2}  {x^6} $
$ \dfrac{{{x^3}}}{5} + \dfrac{2}{5}x + \dfrac{1}{5}  \dfrac{7}{2}{x^2}  {x^6} $
As we know that highest power of variable in a Polynomial is known as degree of a Polynomial. In given Polynomial, the term with highest of x is $  {x^6} $ , and the exponent of x in this term in $ 6 $ .
(ii) the coefficient of $ {x^3} $
Ans: The coefficient of $ {x^3} $ is $ \dfrac{1}{5} $ .
(iii) the coefficient of $ {x^6} $
Ans: The coefficient of $ {x^6} $ is $  1 $ .
(iv) the constant term.
Ans: The constant term is $ \dfrac{1}{5} $ .
4. Write the coefficient of $ {x^2} $ in each of the following:
(i) $ \dfrac{\pi }{6}x + {x^2}  1 $
Ans: Coefficient of $ {x^2} $ in the given Polynomial is $ 1 $ .
(ii) $ 3x  5 $
Ans: The given Polynomial can also be written as $ 0{x^2} + 3x  5 $ . So, the coefficient of $ {x^2} $ in the Polynomial is $ 0 $ .
(iii) $ \left( {x  1} \right)\left( {3x  4} \right) $
Ans: We can write the given polynomial as:
$ \Rightarrow \left( {x  1} \right)\left( {3x  4} \right) $
$ \Rightarrow 3{x^2}  3x  4x + 4 $
$ \Rightarrow 3{x^2}  7x + 4 $
Therefore, the coefficient of $ {x^2} $ in the given Polynomial is $ 3 $ .
(iv) $ \left( {2x  5} \right)\left( {2{x^2}  3x + 1} \right) $
$ \Rightarrow \left( {2x  5} \right)\left( {2{x^2}  3x + 1} \right) $
$ \Rightarrow 4{x^3}  6{x^2} + 2x  10{x^2} + 15x  5 $
$ \Rightarrow 4{x^3}  16{x^2} + 17x  5 $
So, the coefficient of $ {x^2} $ in the given Polynomial is – 16.
5. Classify the following as a constant, linear quadratic and cubic Polynomials:
(i) $ 2  {x^2}  {x^3} $
Ans: To do this question, we have to keep some points in mind:
(A) A Polynomial in which there is no variable term and there is only constant term, is known as a constant polynomial.
(B) A Polynomial having degree 1 is known as a linear Polynomial.
(C) A Polynomial of having degree 2 is known as quadratic Polynomial.
(D) A Polynomial having degree 3 is known as cubic Polynomial.
Cubic Polynomial is the correct answer.
(ii) $ 3{x^3} $
Ans: Cubic Polynomial is the correct answer.
(iii) $ 5t  \sqrt 7 $
Ans: Linear Polynomial is the correct answer.
(iv) $ 4  5{y^2} $
Ans: Quadratic Polynomial is the correct answer.
(v) $ 3 $
Ans: Constant Polynomial is the correct answer.
(vi) $ 2 + x $
(vii) $ {y^3}  y $
(viii) $ 1 + x + {x^3} $
Ans: Cubic Polynomial is the correct answer.
(ix) $ {t^2} $
(x) $ \sqrt 2 x  1 $
6. Give an example of a Polynomial, which is:
(i) monomial of degree $ 1 $
Ans: A Polynomial which contains only one term is called a monomial, a Polynomial having only two terms is called binomial, a Polynomial having only three terms is called a trinomial.
$ 5x $ is monomial of degree $ 1 $ .
(ii) binomial of degree $ 20 $
Ans: $ {x^{20}} + 5 $ is a binomial of degree $ 20 $ .
(iii) trinomial of degree $ 2 $
Ans: $2{x^2} + 2x + 1 $ is a trinomial of degree $ 2 $ .
7. Find the value of the Polynomial $ 3{x^3}  4{x^2} + 7x + 5 $ , when $ x = 3 $ and also when $ x =  3 $ .
Ans: Let p(x) be the given Polynomial.
$ p\left( x \right) = 3{x^2}  4{x^2} + 7x  5 $
Now, put $ x = 3 $
$ p\left( 3 \right) = 3{\left( 3 \right)^3}  4{\left( 3 \right)^2} + 7\left( 3 \right)  5 $
$ p\left( 3 \right) = 3\left( {27} \right)  4\left( 9 \right) + 7\left( 3 \right)  5 $
$ p\left( 3 \right) = 61 $
Put $ x =  3 $
$ p\left( {  3} \right) = 3{\left( {  3} \right)^3}  4{\left( {  3} \right)^2} + 7\left( {  3} \right)  5 $
$ p\left( {  3} \right) = 3\left( {  27} \right)  4\left( 9 \right) + 7\left( {  3} \right)  5 $
$ p\left( {  3} \right) =  81  36  21  5 $
$ p\left( {  3} \right) =  143 $
8. If $ p\left( x \right) = {x^2}  4x + 3 $ , evaluate $ p\left( 2 \right)  p\left( {  1} \right) + p\left( {\dfrac{1}{2}} \right) $
Ans: We have $ p\left( x \right) = {x^2}  4x + 3 $
Now, put $ x = 2 $
$ p\left( 2 \right) = \left( {{2^2}  4 \times 2 + 3} \right) = \left( {4  8 + 3} \right) = \left( {  4 + 3} \right) =  1 $
Now, put $ x =  1 $
$ p\left( {  1} \right) = \left( {{{\left( {  1} \right)}^2}  4 \times \left( {  1} \right) + 3} \right) = \left( {1 + 4 + 3} \right) = 8 $
Now, put $ x = \dfrac{1}{2} $
$ p\left( {\dfrac{1}{2}} \right) = \left( {{{\left( {\dfrac{1}{2}} \right)}^2}  4 \times \dfrac{1}{2} + 3} \right) = \left( {\dfrac{1}{4}  2 + 3} \right) = \left( {1 + \dfrac{1}{4}} \right) = \dfrac{5}{4} $
$ p\left( 2 \right)  p\left( {  1} \right) + p\left( {\dfrac{1}{2}} \right) =  1  8 + \dfrac{5}{4} $
$ p\left( 2 \right)  p\left( {  1} \right) + p\left( {\dfrac{1}{2}} \right) =  9 + \dfrac{5}{4} = \dfrac{{  36 + 5}}{4} =  \dfrac{{31}}{4} $
9. Find $ p\left( 0 \right) $ , $ p\left( 1 \right) $ , $ p\left( {  2} \right) $ for the following Polynomials:
(i) $ p\left( x \right) = 10x  4{x^2}  3 $
Ans: We have
$ p\left( x \right) = 10x  4{x^2}  3 $
Put, $ x = 0 $
$ p\left( 0 \right) = 10\left( 0 \right)  4{\left( 0 \right)^2}  3 =  3 $
Put, $ x = 1 $
$ p\left( 1 \right) = 10\left( 1 \right)  4{\left( 1 \right)^2}  3 = 10  4  3 = 3 $
Put, $ x =  2 $
$ p\left( {  2} \right) = 10\left( {  2} \right)  4{\left( {  2} \right)^2}  3 =  20  16  3 =  39 $
(ii) $ p\left( y \right) = \left( {y + 2} \right)\left( {y  2} \right) $
$ p\left( y \right) = \left( {y + 2} \right)\left( {y  2} \right) $
Put, $ y = 0 $
$ p\left( 0 \right) = \left( {0 + 2} \right)\left( {0  2} \right) =  4 $
Put, $ y = 1 $
$ p\left( 1 \right) = \left( {1 + 2} \right)\left( {1  2} \right) = \left( 3 \right)\left( {  1} \right) =  3 $
Put, $ y =  2 $
$ p\left( {  2} \right) = \left( {  2 + 2} \right)\left( {  2  2} \right) = 0 $
10. Verify whether the following are true or false.
(i) $  3 $ is a zero of $ x  3 $ .
Ans: A number c can be a zero of a Polynomial p(x) if a number c is such that \[p\left( c \right){\text{ }} = {\text{ }}0\].
Let $ p\left( x \right) = x  3 $
$ p\left( {  3} \right) =  3  3 =  6 $
$ p\left( {  3} \right) \ne 0 $
Hence, $  3 $ is not a zero of $ x  3 $ .
So, the result is False.
(ii) $  \dfrac{1}{3} $ is a zero of $ 3x + 1 $ .
Let $ p\left( x \right) = 3x + 1 $
Put, $ x =  \dfrac{1}{3} $
$ p\left( {  \dfrac{1}{3}} \right) = 3\left( {  \dfrac{1}{3}} \right) + 1 $
$ p\left( {  \dfrac{1}{3}} \right) =  1 + 1 = 0 $
Hence, $  \dfrac{1}{3} $ is zero of $ p\left( x \right) = 3x + 1 $ .
So, the result is True.
(iii) $  \dfrac{4}{5} $ is a zero of $ 4  5y $ .
Let $ p\left( y \right) = 4  5(y) $
Put, $ y =  \dfrac{4}{5} $
$ p\left( {  \dfrac{4}{5}} \right) = 4  5\left( {  \dfrac{4}{5}} \right) $
$ p\left( {  \dfrac{4}{5}} \right) = 4 + 4 = 8 $
Hence, $  \dfrac{4}{5} $ is zero of $ p\left( y \right) = 4  5(y) $ .
(iv) $ 0 $ and $ 2 $ are the zeroes of $ {t^2}  2t $
Ans: Let $ p\left( t \right) = {t^2}  2t $
Put, $ t = 0 $
$ p\left( 0 \right) = {\left( 0 \right)^2}  2\left( 0 \right) $
$ p\left( 0 \right) = 0 $
Put, $ t = 2 $
$ p\left( 2 \right) = {\left( 2 \right)^2}  2\left( 2 \right) = 4  4 = 0 $
Hence, $ 0 $ and $ 2 $ are zeroes of the Polynomial $ {t^2}  2t $ .
(v) $  3 $ is a zero $ {y^2} + y  6 $
Ans: Let $ p\left( y \right) = {y^2} + y  6 $
Put, $ y =  3 $
$ p\left( {  3} \right) = {\left( {  3} \right)^2} + \left( {  3} \right)  6 $
$ p\left( {  3} \right) = 9  9 = 0 $
Hence, $  3 $ is a zero of the Polynomial $ {y^2} + y  6 $ .
11. Find the zeroes of the Polynomial in each of the following:
(i) $ p\left( x \right) = x  4 $
Ans: Here we have to Solve the equation $ p\left( x \right) = 0 $ , we get
$ x  4 = 0 $
$ x = 4 $
So, $ 4 $ is a zero of the Polynomial $ x  4 $ .
(ii) $ g\left( x \right) = 3  6x $
Ans: Here we have to Solve the equation $ g\left( x \right) = 0 $ , we get
$ 3  6x = 0 $
$ x = \dfrac{1}{2} $
So, $ \dfrac{1}{2} $ is a zero of the Polynomial $ 3  6x $ .
(iii) $ q\left( x \right) = 2x  7 $
Ans: Here we have to Solve the equation $ q\left( x \right) = 0 $ , we get
$ 2x  7 = 0 $
$ x = \dfrac{7}{2} $
So, $ \dfrac{7}{2} $ is a zero of the Polynomial $ 2x  7 $ .
(iv) $ h\left( y \right) = 2y $
Ans: Here we have to Solve the equation $ h\left( y \right) = 0 $ , we get
$ 2y = 0 $
$ y = 0 $
So, $ 0 $ is a zero of the Polynomial $ 2y $ .
12. Find the zeroes of the Polynomial $ {\left( {x  2} \right)^2}  {\left( {x + 2} \right)^2} $ .
Ans: Let $ p\left( x \right) = {\left( {x  2} \right)^2}  {\left( {x + 2} \right)^2} $
To get the zeroes of p(x), we have to Solve the equation \[p\left( x \right){\text{ }} = {\text{ }}0\].
So, \[p\left( x \right){\text{ }} = {\text{ }}0\]
$ {\left( {x  2} \right)^2}  {\left( {x + 2} \right)^2} = 0 $
$ \left( {\left( {x  2} \right)  \left( {x + 2} \right)} \right)\left( {\left( {x  2} \right) + \left( {x + 2} \right)} \right) = 0 $
\[\left( {  4} \right)\left( {2x} \right) = 0\]
$ x = 0 $
Hence, $ x = 0 $ is the only one zero of $ p\left( x \right) $ .
13. By acute division, find the quotient and the remainder when the first Polynomial is divided by the second $ {x^4} + 1\,;\,x + 1 $ .
Ans: By acute division, we have
14. By remainder Theorem find the remainder, when p(x) is divided by g(x), where
(i) $ p\left( x \right) = {x^3}  2{x^2}  4x  1\,,\,g\left( x \right) = x + 1 $
Ans: According to the remainder theorem, if we divide a polynomial p(x) by a factor \[\left( {{\text{ }}{\mathbf{x}}{\text{ }}{\text{ }}{\mathbf{a}}{\text{ }}} \right)\] ; then you will find a smaller polynomial with a remainder. This remainder which has been obtained is actually a value of p(x) at $ x = a $ , specifically p(a) . So, \[\left( {{\text{ }}{\mathbf{x}}{\text{ }}{\text{ }}{\mathbf{a}}{\text{ }}} \right)\] is the divisor of p(x) if and only if \[p\left( a \right){\text{ }} = {\text{ }}0\] .
We have, $ g\left( x \right) = x + 1 $
Now, put
\[g\left( x \right) = 0\]
$ \Rightarrow x + 1 = 0 $
$ \Rightarrow x =  1 $
Remainder $ = \,p\left( {  1} \right) $
$ = {\left( {  1} \right)^3}  2{\left( {  1} \right)^2}  4\left( {  1} \right) =  1  2 + 4  1 = 0 $
(ii) $ p\left( x \right) = {x^3}  3{x^2} + 4x + 50\,,\,g\left( x \right) = x  3 $
Ans: According to the remainder theorem, if we divide a polynomial p(x) by a factor \[\left( {{\text{ }}{\mathbf{x}}{\text{ }}{\text{ }}{\mathbf{a}}{\text{ }}} \right)\] ; then you will find a smaller polynomial with a remainder. This remainder which has been obtained is actually a value of p(x) at $ x = a $ , specifically p(a) . So, \[\left( {{\text{ }}{\mathbf{x}}{\text{ }}{\text{ }}{\mathbf{a}}{\text{ }}} \right)\] is the divisor of p(x) if and only if \[p\left( a \right){\text{ }} = {\text{ }}0\] .
We have, $ g\left( x \right) = x  3 $
$ \Rightarrow x  3 = 0 $
$ \Rightarrow x = 3 $
Remainder $ = \,p\left( 3 \right) $
$ = {\left( 3 \right)^3}  3{\left( 3 \right)^2} + 4\left( 3 \right) + 50 = 27  27 + 12 + 50 = 62 $
(iii) $ p\left( x \right) = 4{x^3}  12{x^2} + 14x  3\,,\,g\left( x \right) = 2x  1 $
We have, $ g\left( x \right) = 2x  1 $
$ \Rightarrow 2x  1 = 0 $
$ \Rightarrow x = \dfrac{1}{2} $
Remainder $ = \,p\left( {\dfrac{1}{2}} \right) $
$ = 4{\left( {\dfrac{1}{2}} \right)^3}  12{\left( {\dfrac{1}{2}} \right)^2} + 14\left( {\dfrac{1}{2}} \right)  3 = 4\left( {\dfrac{1}{8}} \right)  12\left( {\dfrac{1}{4}} \right) + 7  3 $
$ = \dfrac{1}{2}  3 + 7  3 = \dfrac{1}{2} + 1 = \dfrac{3}{2} $
(iv) $ p\left( x \right) = {x^3}  6{x^2} + 2x  4\,,\,g\left( x \right) = 1  \dfrac{3}{2}x $
We have, $ g\left( x \right) = 1  \dfrac{3}{2}x $
$ \Rightarrow 1  \dfrac{3}{2}x = 0 $
$ \Rightarrow x = \dfrac{2}{3} $
Remainder $ = \,p\left( {\dfrac{2}{3}} \right) $
$ = {\left( {\dfrac{2}{3}} \right)^3}  6{\left( {\dfrac{2}{3}} \right)^2} + 2\left( {\dfrac{2}{3}} \right)  4\, = \dfrac{8}{{27}}  6 \times \dfrac{4}{9} + \dfrac{4}{3}  4 $
$ = \dfrac{8}{{27}}  \dfrac{8}{3} + \dfrac{4}{3}  4 = \dfrac{{8  72 + 36  108}}{{27}} = \dfrac{{  136}}{{27}} $
15. Check whether p(x) is a multiple of g(x) or not:
(i) $ p\left( x \right) = {x^3}  5{x^2} + 4x  3\,,\,g\left( x \right) = x  2 $
p(x) is a multiple of g(x) if g(x) divides p(x)
Now, $ g\left( x \right) = x  2 $ gives $ x = 2 $ if we put $ g\left( x \right) = 0 $ .
Remainder $ = \,p\left( 2 \right) = {\left( 2 \right)^3}  5{\left( 2 \right)^2} + 4\left( 2 \right)  3 $
$ = 8  5\left( 4 \right) + 8  3 = 8  20 + 8  3 =  7 $
Hence, the remainder $ \ne 0 $
Therefore, function p(x) is not a multiple of function g(x).
(ii) $ p\left( x \right) = 2{x^3}  11{x^2}  4x + 5\,,\,g\left( x \right) = 2x + 1 $
p(x) will be a multiple of g(x) if and only if g(x) divides p(x).
$ g\left( x \right) = 2x + 1 $ give $ x =  \dfrac{1}{2} $
remainder $ = p\left( {  \dfrac{1}{2}} \right) = 2{\left( {\dfrac{{  1}}{2}} \right)^3}  11{\left( {  \dfrac{1}{2}} \right)^2}  4\left( {  \dfrac{1}{2}} \right) + 5 $
$ = 2\left( {  \dfrac{1}{8}} \right)  11\left( {\dfrac{1}{4}} \right) + 2 + 5 = \dfrac{{  1}}{4}  \dfrac{{11}}{4} + 7 $
$ = \dfrac{{  1  11 + 28}}{4} = \dfrac{{16}}{4} = 4 $
Since the remainder $ \ne 0 $ .
So, p(x) will not be a multiple of g(x).
16. Show that:
(i) $ x + 3 $ is a factor of $ 69 + 11x  {x^2} + {x^3} $
Ans: Let $ p\left( x \right) = 69 + 11x  {x^2} + {x^3}\,\,,\,\,g\left( x \right) = x + 3 $
$ g\left( x \right) = x + 3 = 0 $
It will give, $ x =  3 $
g(x) will be a factor of p(x) if $ p\left( {  3} \right) = 0 $ (Factor theorem).
Now, $ p\left( {  3} \right) = 69 + 11\left( {  3} \right)  {\left( {  3} \right)^2} + {\left( {  3} \right)^3} $
$ = 69  33  9  27 = 0 $
Hence $ p\left( {  3} \right) = 0 $ , g(x) will be a factor of p(x).
(ii) $ 2x  3 $ is a factor of $ x + 2{x^3}  9{x^2} + 12 $
Ans: Let $ p\left( x \right) = x + 2{x^3}  9{x^2} + 12 $ and $ g\left( x \right) = 2x  3 $
$ g\left( x \right) = 2x  3 $
$ g\left( x \right) = 0 $
$ 2x  3 = 0 $
$ x = \dfrac{3}{2} $ ,
g(x) will be factor of p(x) if $ p\left( {\dfrac{3}{2}} \right) = 0 $ (Factor theorem)
Now, $ p\left( {\dfrac{3}{2}} \right) = \dfrac{3}{2} + 2{\left( {\dfrac{3}{2}} \right)^3}  9{\left( {\dfrac{3}{2}} \right)^2} + 12 = \dfrac{3}{2} + 2\left( {\dfrac{{27}}{8}} \right)  9\left( {\dfrac{9}{4}} \right) + 12 $
$ = \dfrac{3}{2} + \dfrac{{27}}{4}  \dfrac{{81}}{4} + 12 = \dfrac{{6 + 27  81 + 48}}{4} = \dfrac{0}{4} = 0 $
Since, $ p\left( {\dfrac{3}{2}} \right) = 0 $ , therefore, g(x) is a factor of p(x).
17. Determine which of the following Polynomials has $ x  2 $ the factor as:
(i) $ 3{x^2} + 6x  24 $
Ans: As we know that if $ \left( {x  a} \right) $ is a factor of p(x), then p(A) =0.
Let $ p\left( x \right) = 3{x^2} + 6x  24 $
If $ x  2 $ will be factor of $ p\left( x \right) = 3{x^2} + 6x  24 $ , then $ p\left( 2 \right) $ should be equal to $ 0 $ .
Now, $ p\left( 2 \right) = 3{\left( 2 \right)^2} + 6\left( 2 \right)  24 $
$ = 3\left( 4 \right) + 6\left( 2 \right)  24 = 12 + 12  24 = 0 $
Therefore, by factor theorem $ x  2 $ , is factor of $ 3{x^2} + 6x  24 $ .
(ii) $ 4{x^2} + x  2 $
Let $ p\left( x \right) = 4{x^2} + x  2 $
If $ x  2 $ will be factor of $ p\left( x \right) = 4{x^2} + x  2 $ , then $ p\left( 2 \right) $ should be equal to $ 0 $ .
$ p\left( 2 \right) = 4{\left( 2 \right)^2} + 2  2 $
$ = 16 + 2  2 = 16 $
Since, $ 16 \ne 0 $
Therefore, $ x  2 $ is not a factor of $ 4{x^2} + x  2 $ .
18. Show that $ p  1 $ is a factor of $ {p^{10}}  1 $ and also of $ {p^{11}}  1 $ .
Ans: $ p  1 $ is a factor $ {p^{10}}  1 $ , then $ {\left( 1 \right)^{10}}  1 $ should be equal to zero.
Now, $ {\left( 1 \right)^{10}}  1 = 1  1 = 0 $
Therefore, $ p  1 $ is a factor $ {p^{10}}  1 $ .
Again, if $ p  1 $ is a factor of $ {p^{11}}  1 $ , then $ {\left( 1 \right)^{11}}  1 $ should be equal to zero.
Now, $ {\left( 1 \right)^{11}}  1 = 1  1 = 0 $
Therefore, $ p  1 $ is a factor $ {p^{11}}  1 $ .
Hence, $ p  1 $ is a factor of $ {p^{10}}  1 $ and also of $ {p^{11}}  1 $ .
19. For what value of m is $ {x^3}  2m{x^2} + 16 $ divisible by $ x + 2 $ ?
Ans: If $ {x^3}  2m{x^2} + 16 $ is divisible by $ x + 2 $ , then $ x + 2 $ is a factor of $ {x^3}  2m{x^2} + 16 $ .
Now, let $ p\left( x \right) = {x^3}  2m{x^2} + 16 $
As, $ x + 2 = x  \left( {  2} \right) $ is a factor of $ {x^3}  2m{x^2} + 16 $ .
So, $ p\left( {  2} \right) = 0 $
Now, $ p\left( {  2} \right) = {\left( {  2} \right)^3}  2m{\left( {  2} \right)^2} + 16 $
$ =  8  8m + 16 = 8  8m $
Now, $ p\left( {  2} \right) = 0 $
$ \Rightarrow \,8  8m = 0 $
$ \Rightarrow 8m = 8 $
$ \Rightarrow m = 1 $
Hence, for $ m = 1 $ , $ x + 2 $ is a factor of $ {x^3}  2m{x^2} + 16 $ so that $ {x^3}  2m{x^2} + 16 $ is completely divisible by $ x + 2 $ .
20. If $ x + 2a $ is a factor of $ {x^5}  4{a^2}{x^3} + 2x + 2a + 3 $ , find a.
Ans: Let $ p\left( x \right) = {x^5}  4{a^2}{x^3} + 2x + 2a + 3 $
If $ x  \left( {  2a} \right) $ is a factor of $ p(x) $ , then $ p\left( {  2a} \right) = 0 $ .
$ p\left( {  2a} \right) = {\left( {  2a} \right)^5}  4{a^2}{\left( {  2a} \right)^3} + 2\left( {  2a} \right) + 2a + 3 $
$ =  32{a^5} + 32{a^5}  4a + 2a + 3 $
$ =  2a + 3 $
Now, $ p\left( {  2a} \right) = 0 $
$ \Rightarrow  2a + 3 = 0 $
$ \Rightarrow a = \dfrac{3}{2} $
21. Find the value of m so that $ 2x  1 $ be a factor of $ 8{x^4} + 4{x^3}  16{x^2} + 10x + m $ .
Ans: Let $ p\left( x \right) = 8{x^4} + 4{x^3}  16{x^2} + 10x + m $
As, $ \left( {2x  1} \right) $ is a factor of $ p\left( x \right) $ .
Then $ p\left( {\dfrac{1}{2}} \right) = 0 $ (using factor theorem)
$ \Rightarrow 8{\left( {\dfrac{1}{2}} \right)^4} + 4{\left( {\dfrac{1}{2}} \right)^3}  16{\left( {\dfrac{1}{2}} \right)^2} + 10\left( {\dfrac{1}{2}} \right) + m = 0 $
$ \Rightarrow 8\left( {\dfrac{1}{{16}}} \right) + 4\left( {\dfrac{1}{8}} \right)  16\left( {\dfrac{1}{4}} \right) + 5 + m = 0 $
$ \Rightarrow \dfrac{1}{2} + \dfrac{1}{2}  4 + 5 + m = 0 $
$ \Rightarrow 1 + 1 + m = 0 $
$ \Rightarrow \,m =  2 $
22. If $ x + 1 $ is a factor of $ a{x^3} + {x^2}  2x + 4a  9 $ , find the value of a.
Ans: Let $ p\left( x \right) = a{x^3} + {x^2}  2x + 4a  9 $
As $ x + 1 $ is a factor of p(x)
$ p\left( {  1} \right) = 0 $ (By factor theorem)
$ \Rightarrow \,a{\left( {  1} \right)^3} + {\left( {  1} \right)^2}  2\left( {  1} \right) + 4a  9 = 0 $
$ \Rightarrow \,  a + 1 + 2 + 4a  9 = 0 $
$ \Rightarrow \,3a  6 = 0 $
$ \Rightarrow \,3a = 6 $
$ \Rightarrow \,a = 2 $
23. Factorize:
(i) $ {x^2} + 9x + 18 $
Ans: In order to factorize $ {x^2} + 9x + 18 $ , we have to split the middle term.
$ \Rightarrow {x^2} + 9x + 18 $
$ \Rightarrow {x^2} + \left( {6 + 3} \right)x + 18 $
$ \Rightarrow {x^2} + 6x + 3x + 18 $
$ \Rightarrow x\left( {x + 6} \right) + 3\left( {x + 6} \right) $
$ \Rightarrow \left( {x + 3} \right)\left( {x + 6} \right) $
(ii) $ 6{x^2} + 7x  3 $
Ans: In order to factorize $ 6{x^2} + 7x  3 $ , we have to split the middle term.
$ \Rightarrow 6{x^2} + 7x  3 $
$ \Rightarrow 6{x^2} + \left( {9  2} \right)x  3 $
$ \Rightarrow 6{x^2} + 9x  2x  3 $
$ \Rightarrow 3x\left( {2x + 3} \right)  1\left( {2x + 3} \right) $
$ \Rightarrow \left( {3x  1} \right)\left( {2x + 3} \right) $
(iii) $ 2{x^2}  7x  15 $
Ans: If we want to factorise $ 2{x^2}  7x  15 $ , we have to split the middle term.
$ \Rightarrow 2{x^2}  7x  15 $
$ \Rightarrow 2{x^2}  \left( {10  3} \right)x  15 $
$ \Rightarrow 2{x^2}  10x + 3x  15 $
$ \Rightarrow 2x\left( {x  5} \right) + 3\left( {x  5} \right) $
$ \Rightarrow \left( {2x + 3} \right)\left( {x  5} \right) $
(iv) $ 84  2r  2{r^2} $
Ans: If we want to factorise $ 84  2r  2{r^2} $ , we have to split the middle term.
$ \Rightarrow  \left( {2{r^2} + 2r  84} \right) $
$ \Rightarrow  2\left( {{r^2} + r  42} \right) $
$ \Rightarrow  2\left( {{r^2} + \left( {7  6} \right)r  42} \right) $
$ \Rightarrow  2\left( {{r^2} + 7r  6r  42} \right) $
$ \Rightarrow  2\left( {r\left( {r + 7} \right)  6\left( {r + 7} \right)} \right) $
$ \Rightarrow 2\left( {6  r} \right)\left( {r + 7} \right) $
24. Factorise:
(i) $ 2{x^3}  3{x^2}  17x + 30 $
Ans: Let $ f\left( x \right) = 2{x^3}  3{x^2}  17x + 30 $ be the given Polynomial. The factors of the constant term $ + 30 $ are $ \pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30 $ . The factor of coefficient of $ {x^3} $ is $ 2 $ .
Therefore, possible rational roots of f(x) are:
$ \Rightarrow \pm 1, \pm 3, \pm 5, \pm 15, \pm \dfrac{1}{2}, \pm \dfrac{3}{2}, \pm \dfrac{5}{2}, \pm \dfrac{{15}}{2} $
We $ f\left( 2 \right) = 2{\left( 2 \right)^3}  3{\left( 2 \right)^2}  17\left( 2 \right) + 30 $
$ = 2\left( 8 \right)  3\left( 4 \right)  17\left( 2 \right) + 30 $
$ = 16  12  34 + 30 = 0 $
And $ f\left( {  3} \right) = 2{\left( {  3} \right)^3}  3{\left( {  3} \right)^2}  17\left( {  3} \right) + 30 $
$ = 2\left( {  27} \right)  3\left( 9 \right)  17\left( {  3} \right) + 30 $
$ =  54  27 + 51 + 30 = 0 $
Hence, $ \left( {x  2} \right) $ $ \left( {x + 3} \right) $ , will be the factors of $ f\left( x \right) $ .
$ \Rightarrow {x^2} + x  6 $ is a factor of
$ f\left( x \right) $ .
Let's divide $ f\left( x \right) = 2{x^3}  3{x^2}  17x + 30 $ by $ {x^2} + x  6 $ to get the other factors of $ f\left( x \right) $ .
Factors of $ f\left( x \right) $ .
By long division, we have
$ 2{x^3}  3{x^2}  17x + 30 = \left( {{x^2} + x  6} \right)\left( {2x  5} \right) $
$ \Rightarrow \,2{x^3}  3{x^2}  17x + 30 = \left( {x  2} \right)\left( {x + 3} \right)\left( {2x  5} \right) $
Hence, $ \,2{x^3}  3{x^2}  17x + 30 = \left( {x  2} \right)\left( {x + 3} \right)\left( {2x  5} \right) $
(ii) $ {x^3}  6{x^2} + 11x  6 $
Ans: Let $ f\left( x \right) = {x^3}  6{x^2} + 11x  6 $ be the given Polynomial. The factors of the constant term $  6 $ are $ \pm 1, \pm 2, \pm 3\,\,and\,\, \pm 6 $ .
We $ f\left( 1 \right) = {\left( 1 \right)^3}  6{\left( 1 \right)^2} + 11\left( 1 \right)  6 = 0 $
And $ f\left( 2 \right) = {\left( 2 \right)^3}  6{\left( 2 \right)^2} + 11\left( 2 \right)  6 = 8  24 + 22  6 = 0 $
Hence, $ \left( {x  1} \right) $ $ \left( {x  2} \right) $ , will be factors $ f\left( x \right) $ .
$ \Rightarrow \left( {x  1} \right)\left( {x  2} \right) = {x^2}  3x + 2 $ will be a factor of $ f\left( x \right) $ .
Let us divide $ f\left( x \right) = {x^3}  6{x^2} + 11x  6 $ by $ {x^2}  3x + 2 $ to get the other factors of $ f\left( x \right) $ .
$ {x^3}  6{x^2} + 11x  6 = \left( {{x^2}  3x + 2} \right)\left( {x  3} \right) $
$ \Rightarrow {x^3}  6{x^2} + 11x  6 = \left( {x  1} \right)\left( {x  2} \right)\left( {x  3} \right) $
Hence, $ {x^3}  6{x^2} + 11x  6 = \left( {x  1} \right)\left( {x  2} \right)\left( {x  3} \right) $
(iii) $ {x^3} + {x^2}  4x + 4 $
Ans: Let $ f\left( x \right) = {x^3} + {x^2}  4x  4 $ be the given Polynomial. The factors of the constant term $  4 $ are $ \pm 1, \pm 2, \pm 4\, $ .
We $ f\left( {  1} \right) = {\left( {  1} \right)^3} + {\left( {  1} \right)^2}  4\left( {  1} \right)  4 = 0 $
And $ f\left( 2 \right) = {\left( 2 \right)^3} + {\left( 2 \right)^2}  4\left( 2 \right)  4 = 8 + 4  8  4 = 0 $
Hence, $ \left( {x + 1} \right) $ , $ \left( {x  2} \right) $ are factors of $ f\left( x \right) $ .
$ \Rightarrow \left( {x + 1} \right)\left( {x  2} \right) = {x^2}  x  2 $ will be a factor of $ f\left( x \right) $ .
Let's now divide $ f\left( x \right) = {x^3} + {x^2}  4x  4 $ by $ {x^2}  x  2 $ to get the other factors of $ f\left( x \right) $ .
$ {x^3} + {x^2}  4x  4 = \left( {{x^2}  x  2} \right)\left( {x + 2} \right) $
$ \Rightarrow {x^3} + {x^2}  4x  4 = \left( {x + 1} \right)\left( {x  2} \right)\left( {x + 2} \right) $
Hence, $ {x^3} + {x^2}  4x  4 = \left( {x + 1} \right)\left( {x  2} \right)\left( {x + 2} \right) $
(iv) $ 3{x^3}  {x^2}  3x + 1 $
Ans: Let $ f\left( x \right) = 3{x^3}  {x^2}  3x + 1 $ be the given Polynomial. The factors of a constant term $ + 1 $ are $ \pm 1.\, $ The factor of coefficient of $ {x^3} $ is $ 3 $ .Hence, possible rational roots of $ f\left( x \right) $ are $ \pm \dfrac{1}{3} $ .
$ f\left( 1 \right) = 3{\left( 1 \right)^3}  {\left( 1 \right)^2}  3\left( 1 \right) + 1 = 0 $
And $ f\left( {  1} \right) = 3{\left( {  1} \right)^3}  {\left( {  1} \right)^2}  3\left( {  1} \right) + 1 =  3  1 + 3 + 1 = 0 $
Hence, $ \left( {x  1} \right) $ $ \left( {x + 1} \right) $ , will be factors of $ f\left( x \right) $ .
$ \Rightarrow \left( {x  1} \right)\left( {x + 2} \right) = {x^2}  1 $ will be factor of $ f\left( x \right) $ .
Let's now divide $ f\left( x \right) = 3{x^3}  {x^2}  3x + 1 $ by $ {x^2}  1 $ to get the other factors of $ f\left( x \right) $ .
$ 3{x^3}  {x^2}  3x + 1 = \left( {{x^2}  1} \right)\left( {3x  1} \right) $
$ \Rightarrow 3{x^3}  {x^2}  3x + 1 = \left( {x  1} \right)\left( {x + 1} \right)\left( {3x  1} \right) $
Hence, $ 3{x^3}  {x^2}  3x + 1 = \left( {x  1} \right)\left( {x + 1} \right)\left( {3x  1} \right) $
25. Using suitable identify, evaluate the following:
(i) $ {103^3} $
Ans: $ {103^3} = {\left( {100 + 3} \right)^3} $
Now using identify $ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right) $ , we have
$ {\left( {100 + 3} \right)^3} = {\left( {100} \right)^3} + {\left( 3 \right)^3} + 3\left( {100} \right)\left( 3 \right)\left( {100 + 3} \right) $
$ = 1000000 + 27 + 900\left( {100 + 3} \right) $
$ = 1000000 + 27 + 90000 + 2700 $
$ = 1092727 $
(ii) $ 101 \times 102 $
Ans: $ 101 \times 102 $
$ = \left( {100 + 1} \right)\left( {100 + 2} \right) $ (using $ \left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab $ )
$ = {\left( {100} \right)^2} + \left( {1 + 2} \right)100 + \left( 1 \right)\left( 2 \right) $
$ = 10000 + \left( 3 \right)100 + 2 = 10000 + 300 + 2 = 10302 $
(iii) $ {999^2} $
Ans: $ {999^2} $
Ans: $ {\left( {1000  1} \right)^2} $
$ = {\left( {1000} \right)^2}  2\left( {1000} \right) \times 1 + {1^2} $
$ = 1000000  2000 + 1 = 998001 $
26. Factorise the following:
(i) $ 4{x^2} + 20x + 25 $
Ans: We have,
$ 4{x^2} + 20x + 25 = {\left( {2x} \right)^2} + 2\left( {2x} \right)5 + {\left( 5 \right)^2} $
$ = {\left( {2x + 5} \right)^2} $ $ \left[ {{a^2} + 2ab + {b^2} = {{\left( {a + b} \right)}^2}} \right] $
$ = \left( {2x + 5} \right)\left( {2x + 5} \right) $
(ii) $ 9{y^2}  66yz + 121{z^2} $
$ 9{y^2}  66yz + 121{z^2} = {\left( {3y} \right)^2}  2\left( {3y} \right)\left( {11z} \right) + {\left( {  11z} \right)^2} $
$ = {\left( {3y  11z} \right)^2} $ $ \left[ {{a^2} + {b^2}  2ab = {{\left( {a  b} \right)}^2}} \right] $
$ = \left( {3y  11} \right)\left( {3y  11} \right) $
(iii) $ {\left( {2x + \dfrac{1}{3}} \right)^2}  {\left( {x  \dfrac{1}{2}} \right)^2} $
Ans: $ {\left( {2x + \dfrac{1}{3}} \right)^2}  {\left( {x  \dfrac{1}{2}} \right)^2} $
Now, using identity $ {a^2}  {b^2} = \left( {a + b} \right)\left( {a  b} \right) $
$ = \left[ {\left( {2x + \dfrac{1}{3}} \right) + \left( {x  \dfrac{1}{2}} \right)} \right]\left[ {\left( {2x + \dfrac{1}{3}} \right)  \left( {x  \dfrac{1}{2}} \right)} \right] $
$ = \left( {2x + \dfrac{1}{3} + x  \dfrac{1}{2}} \right)\left( {2x + \dfrac{1}{3}  x + \dfrac{1}{2}} \right) = \left( {3x  \dfrac{1}{6}} \right)\left( {x + \dfrac{5}{6}} \right) $
27. Factorise the following:
(i) $ 9{x^2}  12x + 3 $
Ans: we have,
$ 9{x^2}  12x + 3 = 9{x^2}  9x  3x + 3 $
$ = 9x\left( {x  1} \right)  3\left( {x  1} \right) $
$ = \left( {9x  3} \right)\left( {x  1} \right) $
$ = 3\left( {3x  1} \right)\left( {x  1} \right) $
(ii) $ 9{x^2}  12x + 4 $
$ 9{x^2}  12x + 4 = {\left( {3x} \right)^2}  2\left( {3x} \right)\left( 2 \right) + {\left( 2 \right)^2} $
$ = {\left( {3x  2} \right)^2}\left[ {{a^2}  2ab + {b^2} = {{\left( {a  b} \right)}^2}} \right] $
$ = \left( {3x  2} \right)\left( {3x  2} \right) $
28. Expand the following:
(i) $ {\left( {4a  b + 2c} \right)^2} $
$ {\left( {4a  b + 2c} \right)^2} = {\left( {4a} \right)^2} + {\left( {  b} \right)^2} + {\left( {2c} \right)^2} + 2\left( {4a} \right)\left( {  b} \right) + 2\left( {  b} \right)\left( {2c} \right) + 2\left( {2c} \right)\left( {4a} \right) $
Using $ \left[ {{a^2} + b + {c^2} + 2ab + 2bc + 2ca = {{\left( {a + b + c} \right)}^2}} \right] $
$ = 16{a^2} + {b^2} + 4{c^2}  8ab  4bc + 16ca $
(ii) $ {\left( {3a  5b  c} \right)^2} $
$ {\left( {3a  5b  c} \right)^2} = {\left( {3a} \right)^2} + {\left( {  5b} \right)^2} + {\left( {  c} \right)^2} + 2\left( {3a} \right)\left( {  5b} \right) + 2\left( {  5b} \right)\left( {  c} \right) + 2\left( {  c} \right)\left( {3a} \right) $
Using $ \left[ {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = {{\left( {a + b + c} \right)}^2}} \right] $
$ = 9{a^2} + 25{b^2} + {c^2}  30ab + 10bc  6ca $
(iii) $ {\left( {  x + 2y  3z} \right)^2} $
$ {\left( {  x + 2y  3z} \right)^2} = {\left( {  x} \right)^2} + {\left( {2y} \right)^2} + {\left( {  3z} \right)^2} + 2\left( {  x} \right)\left( {2y} \right) + 2\left( {2y} \right)\left( {  3z} \right) + 2\left( {  3z} \right)\left( {  x} \right) $
$ = {x^2} + 4{y^2} + 9{z^2}  4xy  12yz + 6xz $
29. Factorise the following:
(i) $ 9{x^2} + 4{y^2} + 16{z^2} + 12xy  16yz  24xz $
$ \Rightarrow {\left( {3x} \right)^2} + {\left( {2y} \right)^2} + {\left( {  4z} \right)^2} + 2\left( {3x} \right)\left( {2y} \right) + 2\left( {2y} \right)\left( {  4z} \right) + 2\left( {  4z} \right)\left( {3x} \right) $
$ \Rightarrow {\left\{ {3x + 2y + \left( {  4z} \right)} \right\}^2}\,\left[ {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = {{\left( {a + b + c} \right)}^2}} \right] $
$ \Rightarrow {\left( {3x + 2y  4z} \right)^2} = \left( {3x + 2y  4z} \right)\left( {3x + 2y  4z} \right) $
(ii) $ 25{x^2} + 16{y^2} + 4{z^2}  40xy + 16yz  20xz $
$ \Rightarrow {\left( {  5x} \right)^2} + {\left( {4y} \right)^2} + {\left( {2z} \right)^2} + 2\left( {  5x} \right)\left( {4y} \right) + 2\left( {4y} \right)\left( {2z} \right) + 2\left( {2z} \right)\left( {  5x} \right) $
$ \Rightarrow {\left\{ {  5x + 4y + 2z} \right\}^2}\,\left[ {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = {{\left( {a + b + c} \right)}^2}} \right] $
(iii) $ 16{x^2} + 4{y^2} + 9{z^2}  16xy  12yz + 24xz $
$ \Rightarrow {\left( {4x} \right)^2} + {\left( {  2y} \right)^2} + {\left( {3z} \right)^2} + 2\left( {4x} \right)\left( {  2y} \right) + 2\left( {  2y} \right)\left( {3z} \right) + 2\left( {3z} \right)\left( {4x} \right) $
$ \Rightarrow {\left\{ {4x  2y + 3z} \right\}^2}\,\left[ {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = {{\left( {a + b + c} \right)}^2}} \right] $
$ \Rightarrow {\left( {4x  2y + 3z} \right)^2} = \left( {4x  2y + 3z} \right)\left( {4x  2y + 3z} \right) $
30. If $ a + b + c = 9 $ and $ ab + bc + ca = 26 $ , find $ {a^2} + {b^2} + {c^2} $ .
Ans: We know that
$ {\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca $
$ \Rightarrow {\left( {a + b + c} \right)^2} = \left( {{a^2} + {b^2} + {c^2}} \right) + 2\left( {ab + bc + ca} \right) $
Now, on putting the values, we get
$ \Rightarrow {\left( 9 \right)^2} = \left( {{a^2} + {b^2} + {c^2}} \right) + 2\left( {26} \right) $
$ \Rightarrow {a^2} + {b^2} + {c^2} = 81  52 = 29 $
31. Expand the following:
(i) $ {\left( {3a  2b} \right)^3} $
$ {\left( {3a  2b} \right)^3} = {\left( {3a} \right)^3}  {\left( {2b} \right)^3}  3\left( {3a} \right)\left( {2b} \right)\left( {3a  2b} \right) $
Now, using identity $ \left[ {{{\left( {a  b} \right)}^3} = {a^3}  {b^3}  3ab\left( {a  b} \right)} \right] $
$ = 27{a^3}  8{b^3}  54{a^2}b + 36a{b^2} $
(ii) $ {\left( {\dfrac{1}{x} + \dfrac{y}{3}} \right)^3} $
Ans: Here, using the identity $ \left[ {{{\left( {a + b} \right)}^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)} \right] $
$ {\left( {\dfrac{1}{x} + \dfrac{y}{3}} \right)^3} = {\left( {\dfrac{1}{x}} \right)^3} + {\left( {\dfrac{y}{3}} \right)^3} + 3 \times \dfrac{1}{x} \times \dfrac{y}{3}\left( {\dfrac{1}{x} + \dfrac{y}{3}} \right) $
$ \Rightarrow \dfrac{1}{{{x^3}}} + \dfrac{{{y^3}}}{{27}} + \dfrac{y}{x}\left( {\dfrac{1}{x} + \dfrac{y}{3}} \right) $
$ \Rightarrow \dfrac{1}{{{x^3}}} + \dfrac{{{y^3}}}{{27}} + \dfrac{y}{{{x^2}}} + \dfrac{{{y^2}}}{{3x}} $
(iii) $ {\left( {4  \dfrac{1}{{3x}}} \right)^3} $
$ {\left( {4  \dfrac{1}{{3x}}} \right)^3} = {\left( 4 \right)^3}  {\left( {\dfrac{1}{{3x}}} \right)^3}  3\left( 4 \right)\left( {\dfrac{1}{{3x}}} \right)\left( {4  \dfrac{1}{{3x}}} \right) $
$ = 64  \dfrac{1}{{27{x^3}}}  \dfrac{4}{x}\left( {4  \dfrac{1}{{3x}}} \right) $
$ = 64  \dfrac{1}{{27{x^3}}}  \dfrac{{16}}{x} + \dfrac{4}{{3{x^2}}} $
32.Factorise the following:
(i) $ 1  64{a^3}  12a + 48{a^2} $
$ 1  64{a^3}  12a + 48{a^2} = {\left( 1 \right)^3}  {\left( {4a} \right)^3}  3\left( 1 \right)\left( {4a} \right)\left( {1  4a} \right) $
$ = {\left( {1  4a} \right)^3}\left[ {{a^3}  {b^3}  3ab\left( {a  b} \right) = {{\left( {a  b} \right)}^3}} \right] $
$ = \left( {1  4a} \right)\left( {1  4a} \right)\left( {1  4a} \right) $
(ii) $ 8{p^3} + \dfrac{{12}}{5}{p^2} + \dfrac{6}{{25}}p + \dfrac{1}{{125}} $
$ 8{p^3} + \dfrac{{12}}{5}{p^2} + \dfrac{6}{{25}}p + \dfrac{1}{{125}} $
$ = {\left( {2p} \right)^3} + 3 \times {\left( {2p} \right)^2} \times \dfrac{1}{5} + 3 \times \left( {2p} \right) \times {\left( {\dfrac{1}{5}} \right)^2} + {\left( {\dfrac{1}{5}} \right)^3} $
$ = {\left( {2p} \right)^3} + {\left( {\dfrac{1}{5}} \right)^3} + 3 \times \left( {2p} \right) \times \dfrac{1}{5}\left[ {2p + \dfrac{1}{5}} \right] $
$ = {\left( {2p + \dfrac{1}{5}} \right)^3} = \left( {2p + \dfrac{1}{5}} \right)\left( {2p + \dfrac{1}{5}} \right)\left( {2p + \dfrac{1}{5}} \right) $
33. Find the following products:
(i) $ \left( {\dfrac{x}{2} + 2y} \right)\left( {\dfrac{{{x^2}}}{4}  xy + 4{y^2}} \right) $
$ \left( {\dfrac{x}{2} + 2y} \right)\left( {\dfrac{{{x^2}}}{4}  xy + 4{y^2}} \right) = \left( {\dfrac{x}{y} + 2y} \right)\left( {{{\left( {\dfrac{x}{2}} \right)}^2}  \left( {\dfrac{x}{2}} \right)\left( {2y} \right) + {{\left( {2y} \right)}^2}} \right) $
$ = {\left( {\dfrac{x}{2}} \right)^3} + {\left( {2y} \right)^3}\left[ {\left( {a + b} \right)\left( {{a^2}  ab + {b^2}} \right) = {a^3} + {b^3}} \right] $
$ = \dfrac{{{x^3}}}{8} + 8{y^3} $
(ii) $ \left( {{x^2}  1} \right)\left( {{x^4} + {x^2} + 1} \right) $
$ \left( {{{\left( x \right)}^2}  1} \right)\left( {{x^4} + {x^2} + 1} \right) = \left( {{x^2}  1} \right)\left( {{{\left( {{x^2}} \right)}^2} + \left( {{x^2}} \right)\left( 1 \right) + {{\left( 1 \right)}^2}} \right) $
$ = {\left( {{x^2}} \right)^3}  {\left( 1 \right)^3}\left[ {\left( {a  b} \right)\left( {{a^2} + ab + {b^2}} \right) = {a^3}  {b^3}} \right] $
$ = {x^6}  1 $
34. Factorise:
(i) $ 1 + 64{x^3} $
$ 1 + 64{x^3} = {\left( 1 \right)^3} + {\left( {4x} \right)^3} $
$ = \left( {1 + 4x} \right)\left( {{{\left( 1 \right)}^2}  \left( 1 \right)\left( {4x} \right) + {{\left( {4x} \right)}^2}} \right) $
$ = \left( {1 + 4x} \right)\left( {1  4x + 16{x^2}} \right) $
(ii) $ {a^3}  2\sqrt {2{b^3}} $
$ {a^3}  2\sqrt {2{b^3}} = {\left( a \right)^3}  {\left( {\sqrt 2 b} \right)^3} $
$ = \left( {a  \sqrt 2 b} \right)\left( {{{\left( a \right)}^2} + \left( a \right)\left( {\sqrt 2 b} \right) + {{\left( {\sqrt 2 b} \right)}^2}} \right) $
$ = \left( {a  \sqrt 2 b} \right)\left( {{a^2} + \sqrt 2 ab + 2{b^2}} \right) $
35. Find the following product:
$ \left( {2x  y + 3z} \right)\left( {4{x^2} + {y^2} + 9{z^2} + 2xy + 3yz  6xz} \right) $
$ = \left\{ {2x + \left( {  y} \right) + 3z} \right\}\left\{ {{{\left( {2x} \right)}^2} + {{\left( {  y} \right)}^2} + {{\left( {3z} \right)}^2}  2x\left( {  y} \right)  \left( {  y} \right)\left( {3z} \right)  \left( {  y} \right)\left( {3z} \right)  \left( {3z} \right)\left( {2x} \right)} \right\} $
$ = {\left( {2x} \right)^3} + {\left( {  y} \right)^3} + {\left( {3z} \right)^3}  3\left( {2x} \right)\left( {  y} \right)\left( {3z} \right) $
using identity $ \left[ {\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2}  ab  bc  ca} \right) = {a^3} + {b^3} + {c^3}  3abc} \right] $
$ = 8{x^3}  {y^3} + 27{z^2} + 18xyz $
36.Factorise:
(i) $ {a^3}  8{b^3}  64{c^3}  24abc $
$ {a^3}  8{b^3}  64{c^3}  24abc $
$ = \left\{ {{{\left( a \right)}^3} + {{\left( {  2b} \right)}^3} + {{\left( {  4c} \right)}^3}  3\left( a \right)\left( {  2b} \right)\left( {  4c} \right)} \right\} $
$ = \left\{ {a + \left( {  2b} \right) + \left( {  4c} \right)} \right\}\left\{ {{a^2} + {{\left( {  2b} \right)}^2} + {{\left( {  4c} \right)}^2}  a\left( {  2b} \right)  \left( {  2b} \right)\left( {  4c} \right)  \left( {  4c} \right)a} \right\} $
Using identity $ \left[ {\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2}  ab  bc  ca} \right) = {a^3} + {b^3} + {c^3}  3abc} \right] $
$ = \left( {a  2b  4c} \right)\left( {{a^2} + 4{b^2} + 16{c^2} + 2ab  8bc + 4ca} \right) $
(ii) $ 2\sqrt 2 {a^3} + 8{b^3}  27{c^3} + 18\sqrt 2 abc $
$ 2\sqrt 2 {a^3} + 8{b^3}  27{c^3} + 18\sqrt 2 abc $
$ = \left\{ {{{\left( {\sqrt 2 a} \right)}^3} + {{\left( {2b} \right)}^3} + {{\left( {  3c} \right)}^3}  3\left( {\sqrt 2 a} \right)\left( {2b} \right)\left( {  3c} \right)} \right\} $
$ = \left\{ {\sqrt 2 a + 2b + \left( {  3c} \right)} \right\}\left\{ {{{\left( {\sqrt 2 a} \right)}^2} + {{\left( {2b} \right)}^2} + {{\left( {  3c} \right)}^2}  \left( {\sqrt 2 a} \right)\left( {2b} \right)  \left( {2b} \right)\left( {  3c} \right)  \left( {  3c} \right)\left( {\sqrt 2 a} \right)} \right\} $
$ = \left( {\sqrt 2 a + 2b  3c} \right)\left( {2{a^2} + 4{b^2} + 9{c^2}  2\sqrt 2 ab + 6bc + 3\sqrt 2 ca} \right) $
37. Without actually calculating the cubes, find the value of:
(i) $ {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{3}} \right)^3}  {\left( {\dfrac{5}{6}} \right)^3} $
Ans: Let $ a = \dfrac{1}{2}\,,\,b = \dfrac{1}{3}\,,\,c =  \dfrac{5}{6} $
$ a + b + c = \dfrac{1}{2} + \dfrac{1}{3}  \dfrac{5}{6} $
$ = \dfrac{{3 + 2  5}}{6} = \dfrac{0}{6} = 0 $
$ \left[ {\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2}  ab  bc  ca} \right) = {a^3} + {b^3} + {c^3}  3abc} \right] $
$ \Rightarrow 3 \times \dfrac{1}{2} \times \dfrac{1}{3} \times  \dfrac{5}{6} $
$ \Rightarrow  \dfrac{5}{{12}} $
(ii) $ {\left( {0.2} \right)^3}  {\left( {0.3} \right)^3} + {\left( {0.1} \right)^3} $
$ {\left( {0.2} \right)^3}  {\left( {0.3} \right)^3} + {\left( {0.1} \right)^3} = {\left( {0.2} \right)^3} + {\left( {  0.3} \right)^3} + {\left( {0.1} \right)^3} $
$ a = 0.2\,,\,b =  0.3\,,\,c = 0.1 $ . Then,
$ a + b + c = 0.2  0.3  0.1 = 0 $ .
$ \Rightarrow 3 \times 0.2 \times \left( {  0.3} \right) \times \left( {0.1} \right) $
$ \Rightarrow  0.018 $
Hence, $ {\left( {0.2} \right)^3}  {\left( {0.3} \right)^3} + {\left( {0.1} \right)^3} =  0.018 $
38. Without finding the cubes, factorise
$ {\left( {x  2y} \right)^3} + {\left( {2y  3z} \right)^3} + {\left( {3z  x} \right)^3} $
$ a = x  2y\,,\,b = 2y  3z\,,\,c = 3z  x $ . Then,
$ a + b + c = x  2y + 2y  3z + 3z  x = 0 $ .
$ \Rightarrow 3 \times \left( {x  2y} \right) \times \left( {2y  3z} \right) \times \left( {3z  x} \right) $
$ \Rightarrow 3\left( {x  2y} \right)\left( {2y  3z} \right)\left( {3z  x} \right) $
39. Find the value of
(i) $ {x^3} + {y^3}  12xy + 64 $ , when $ x + y =  4 $
Ans: $ {x^3} + {y^3}  12xy + 64 = {x^3} + {y^3} + {4^3}  3xy\left( 4 \right) $
$ = \left( {x + y + 4} \right)\left( {{x^2} + {y^2} + {4^2}  xy  4y  4x} \right) $
$ \left[ {x + y =  4} \right] $
$ = \left( 0 \right)\left( {{x^2} + {y^2} + {4^2}  xy  4y  4x} \right) = 0 $
(ii) $ {x^3}  8{y^3}  36xy  216 $ , when $ x = 2y + 6 $
Ans: $ {x^3}  8{y^3}  36xy  216 = {x^3} + {\left( {  2y} \right)^3} + {\left( {  6} \right)^3}  3x\left( {  2y} \right)\left( {  6} \right) $
$ = \left( {x  2y  6} \right)\left( {{x^2} + {{\left( {  2y} \right)}^2} + {{\left( {  6} \right)}^2}  x\left( {  2y} \right)  \left( {  2y} \right)\left( {  6} \right)  x\left( {  6} \right)} \right) $
$ = \left( {x  \left( {2y + 6} \right)} \right)\left( {{x^2} + {{\left( {  2y} \right)}^2} + {{\left( {  6} \right)}^2}  x\left( {  2y} \right)  \left( {  2y} \right)\left( {  6} \right)  x\left( {  6} \right)} \right) $
$ \left[ {x = 2y + 6} \right] $
$ = \left( 0 \right)\left( {{x^2} + 4{y^2} + 36 + 2xy  12y + 6x} \right) = 0 $
40. Give possible experiments for the length and breadth of the rectangle whose area is given by $ 4{a^2} + 4a  3 $
Ans: Area $ 4{a^2} + 4a  3 $ .
Here we will use a method of splitting the middle term.
Here we have to split the middle term as $ 4a = 6a  2a $
$ \Rightarrow 4{a^2} + \left( {6a  2a} \right)  3 $
$ \Rightarrow 4{a^2} + 6a  2a  3 $
$ \Rightarrow 2a\left( {2a + 3} \right)  1\left( {2a + 3} \right) $
$ \Rightarrow \left( {2a  1} \right)\left( {2a + 3} \right) $
We know that, area of rectangle \[ = 4{a^2} + 4a  3\]
Here we also know that, area of a rectangle $ = \,length\, \times \,breadth $ and $ 4{a^2} + 4a  3 = \left( {2a  1} \right)\left( {2a + 3} \right) $ .
Therefore, its possible length and breadth $ = \,\left( {2a  1} \right) $ and $ \left( {2a + 3} \right) $ or, we can say that $ length = \left( {2a + 3} \right) $ and $ breadth\, = \,\left( {2a  1} \right) $ .
Long Answer Questions
Sample Question 1. If \[x{\text{ }} + {\text{ }}y{\text{ }} = {\text{ }}12\] and \[xy{\text{ }} = {\text{ }}27\], find the value of \[{x^3}{\text{ }} + {\text{ }}{y^3}\].
Ans: Here we will use an identity, \[{\text{ }}{\left( {x{\text{ }} + {\text{ }}y} \right)^3}{\text{ = }}{x^3}{\text{ }} + {\text{ }}{y^3}{\text{ + }}3xy{\text{ }}\left( {x{\text{ }} + {\text{ }}y} \right)\]
\[{x^3}{\text{ }} + {\text{ }}{y^3}{\text{ }} = {\text{ }}{\left( {x{\text{ }} + {\text{ }}y} \right)^3}{\text{ }}{\text{ }}3xy{\text{ }}\left( {x{\text{ }} + {\text{ }}y} \right)\]
\[{\text{ = }}{12^3}{\text{ }}{\text{ }}3{\text{ }} \times {\text{ }}27{\text{ }} \times {\text{ }}12\]
\[ = {\text{ }}12{\text{ }}\left[ {{{12}^2}{\text{ }}{\text{ }}3{\text{ }} \times {\text{ }}27} \right]\]
\[ = {\text{ }}12{\text{ }} \times {\text{ }}63\]
\[ = {\text{ }}756\]
Therefore, the value of \[{x^3}{\text{ }} + {\text{ }}{y^3} = 756\] .
EXERCISE 2.4
1. If the Polynomials $ a{z^3} + 4{z^2} + 3z  4 $ and $ {z^3}  4z + a $ leave the same remainder when divided $ z  3 $ , by Find the value of a.
$ p\left( z \right) = a{z^3} + 4{z^2} + 3z  4 $
And $ q\left( z \right) = {z^3}  4z + a $
As it is given that both the Polynomials leave the same remainder when divided by $ z  3 $ .
$ p\left( 3 \right) = q\left( 3 \right) $
\[ \Rightarrow a{\left( 3 \right)^3} + 4{\left( 3 \right)^2} + 3\left( 3 \right)  4 = \,{\left( 3 \right)^3}  4\left( 3 \right) + a\]
\[ \Rightarrow 27a + 4 \times 9 + 9  4 = \,27  12 + a\]
\[ \Rightarrow 27a + 36 + 5 = \,15 + a\]
$ \Rightarrow 27a  a = 15  41 $
$ \Rightarrow 26a =  26 $
$ \Rightarrow a =  1 $
Hence, the required value of a is $  1 $ .
2. The Polynomial $ p\left( x \right) = {x^4}  2{x^3} + 3{x^2}  ax + 3a  7 $ when divided by $ x + 1 $ leave remainder $ 19 $ . Also, find the remainder when $ p\left( x \right) $ is divided by $ x + 2 $ .
Ans: We know that when $ p\left( x \right) $ is divided by $ x + b $ , then the remainder $ = \,p\left( {  b} \right) $
Now, $ p\left( x \right) = {x^4}  2{x^3} + 3{x^2}  ax + 3a  7 $ is divided by $ x + 1 $ , then the remainder $ = \,p\left( {  1} \right) $ .
Also, we know that the remainder is $ 19 $ .
$ \,p\left( {  1} \right) = 19 $
$ {\left( {  1} \right)^4}  2{\left( {  1} \right)^3} + 3{\left( {  1} \right)^2}  a\left( {  1} \right) + 3a  7 = 19 $
$ \Rightarrow 1 + 2 + 3 + a + 3a  7 = 19 $
$ \Rightarrow 4a  1 = 19 $
$ \Rightarrow 4a = 20 $
$ \Rightarrow a = 5 $
Now, $ p\left( x \right) = {x^4}  2{x^3} + 3{x^2}  5x + 3\left( 5 \right)  7 = {x^4}  2{x^3} + 3{x^2}  5x + 8 $
Again, when p(x) is divided by $ x + 2 $ , then
Remainder $ = p\left( {  2} \right) = {\left( {  2} \right)^4}  2{\left( {  2} \right)^3} + 3{\left( {  2} \right)^2}  5\left( {  2} \right) + 8 $
$ = 16 + 16 + 12 + 10 + 8 $
$ = 62 $
3. If both \[\left( {x  2} \right)\] and $ \left( {x  \dfrac{1}{2}} \right) $ are factors of $ p{x^2} + 5x + r $ , Show that $ p = r $ .
Ans: Let $ q\left( x \right) = p{x^2} + 5x + r $
As $ \left( {x  2} \right) $ is a factor of $ q\left( x \right) $
Then, $ q\left( 2 \right) = 0 $
$ \Rightarrow p{\left( 2 \right)^2} + 5\left( 2 \right) + r = 0 $
$ \Rightarrow 4p + 10 + r = 0 $
$ \Rightarrow  4p  r = 10............\left( 1 \right) $
Again, $ \left( {x  \dfrac{1}{2}} \right) $ is a factor of $ q\left( x \right) $ .
Then, $ q\left( {\dfrac{1}{2}} \right) = 0 $
$ \Rightarrow p{\left( {\dfrac{1}{2}} \right)^2} + 5\left( {\dfrac{1}{2}} \right) + r = 0 $
$ \Rightarrow \dfrac{p}{4} + \dfrac{5}{2} + r = 0 $
$ \Rightarrow p + 10 + 4r = 0 $
Now, using equation $ 1 $
$ \Rightarrow p  4p  r + 4r = 0 $
$ \Rightarrow  3p + 3r = 0 $
$ \Rightarrow p = r $
Hence, proved.
4. Without actual division, prove that $ 2{x^4}  5{x^3} + 2{x^2}  x + 2 $ is divisible by $ {x^2}  3x + 2 $ .
$ {x^2}  3x + 2 $
We can also write it as
$ \Rightarrow {x^2}  3x + 2 = {x^2}  2x  x + 2 $
$ \Rightarrow x\left( {x  2} \right)  1\left( {x  2} \right) = \left( {x  1} \right)\left( {x  2} \right) $
Now, we have
$ p\left( x \right) = 2{x^4}  5{x^3} + 2{x^2}  x + 2 $
Put $ x = 1 $
$ p\left( 1 \right) = 2{\left( 1 \right)^4}  5{\left( 1 \right)^3} + 2{\left( 1 \right)^2}  \left( 1 \right) + 2 $
$ p\left( 1 \right) = 2  5 + 2 + 1 = 0 $
$ p\left( 1 \right) = 0 $
Therefore, $ x  1 $ divides p(x).
Put $ x = 2 $
$ p\left( 2 \right) = 2{\left( 2 \right)^4}  5{\left( 2 \right)^3} + 2{\left( 2 \right)^2}  \left( 2 \right) + 2 $
$ p\left( 2 \right) = 32  40 + 8  2 + 2 = 0 $
$ p\left( 2 \right) = 0 $
Therefore, $ x  2 $ divides p(x).
Hence, $ {x^2}  3x + 2 $ divides $ 2{x^4}  5{x^3} + 2{x^2}  x + 2 $ .
5. Simplify $ {\left( {2x  5y} \right)^3}  {\left( {2x + 5y} \right)^3} $ .
$ = {\left( {2x  5y} \right)^3}  {\left( {2x + 5y} \right)^3} $
Now, using $ {a^3}  {b^3} = \left( {a  b} \right)\left( {{a^2} + ab + {b^2}} \right) $
$ = \left\{ {\left( {2x  5y} \right)  \left( {2x + 5y} \right)} \right\}\left\{ {{{\left( {2x  5y} \right)}^2} + \left( {2x  5y} \right)\left( {2x + 5y} \right) + {{\left( {2x + 5} \right)}^2}} \right\} $
$ = \left( {  10y} \right)\left( {2{x^2} + 25{y^2}} \right) $
$ =  120{x^2}y  250{y^3} $
6. Multiply $ {x^2} + 4{y^2} + {z^2} + 2xy + xz  2yz $ by $ \left( {  z + x  2y} \right) $ .
$ \left( {  z + x  2y} \right)\left( {{x^2} + 4{y^2} + {z^2} + 2xy + xz  2yz} \right) $
$ = \left\{ {x + \left( {  2y} \right) + \left( {  z} \right)} \right\}\left\{ {{{\left( x \right)}^2} + {{\left( {  2y} \right)}^2} + {{\left( {  z} \right)}^2}  \left( x \right)\left( {  2y} \right)  \left( x \right)\left( {  z} \right)  \left( 2y \right)\left( {  z} \right)} \right\} $
$ = {x^3} + {\left( {  2y} \right)^3} + {\left( {  z} \right)^3}  3\left( x \right)\left( {  2y} \right)\left( {  z} \right) $
$ = {x^3}  8{y^3}  {z^3}  6xyz $
7. If a, b, c are all nonzero and \[a + b + c = 0\], prove that $ \dfrac{{{a^2}}}{{bc}} + \dfrac{{{b^2}}}{{ca}} + \dfrac{{{c^2}}}{{ab}} = 3 $ .
Ans: Here, a, b, c are all nonzero and \[a + b + c = 0\].
$ {a^3} + {b^3} + {c^3} = 3abc $
Now, we will divide the complete equation by abc.
$ \dfrac{{{a^3}}}{{abc}} + \dfrac{{{b^3}}}{{abc}} + \dfrac{{{c^3}}}{{abc}} = \dfrac{{3abc}}{{abc}} $
$ \dfrac{{{a^2}}}{{bc}} + \dfrac{{{b^2}}}{{ca}} + \dfrac{{{c^2}}}{{ab}} = 3 $ .
8. If $ a + b + c = 5 $ and $ ab + bc + ca = 10 $ , then prove that $ {a^3} + {b^3} + {c^3}  3abc =  25 $ .
Ans: We know that,
$ = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2}  \left( {ab + bc + ca} \right)} \right) $
It is given that $ a + b + c = 5 $ and $ ab + bc + ca = 10 $ .
$ = 5\left( {{a^2} + {b^2} + {c^2}  10} \right) $
Now, we have
$ a + b + c = 5 $
Squaring both sides, we get
$ {\left( {a + b + c} \right)^2} = {\left( 5 \right)^2} $
$ {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) = 25 $
$ {a^2} + {b^2} + {c^2} + 2\left( {10} \right) = 25 $
$ {a^2} + {b^2} + {c^2} = 25  20 = 5 $
Now, put the above value in equation $ = 5\left( {{a^2} + {b^2} + {c^2}  10} \right) $
$ = 5\left( {5  10} \right) $
$ = 5 \times  5 =  25 $
9. Prove that $ {\left( {a + b + c} \right)^3}  {a^3}  {b^3}  {c^3} = 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) $ .
Ans: $ {\left( {a + b + c} \right)^3} = {\left[ {a + \left( {b + c} \right)} \right]^3} $
$ = {a^3} + 3{a^2}\left( {b + c} \right) + 3a{\left( {b + c} \right)^2} + {\left( {b + c} \right)^3} $
$ = {a^3} + 3{a^2}b + 3{a^2}c + 3a\left( {{b^2} + 2bc + {c^2}} \right) + \left( {{b^3} + 3{b^2}c + 3b{c^2} + {c^3}} \right) $
$ = {a^3} + 3{a^2}b + 3{a^2}c + 3a{b^2} + 6abc + 3a{c^2} + {b^3} + 3{b^2}c + 3b{c^2} + {c^3} $
$ = {a^3} + {b^3} + {c^3} + 3{a^2}b + 3{a^2}c + 3a{b^2} + 6abc + 3a{c^2} + 3{b^2}c + 3b{c^2} $
$ = {a^3} + {b^3} + {c^3} + 3abc + 3{a^2}b + 3a{c^2} + 3{a^2}c + 3{b^2}c + 3{b^2}a + 3b{c^2} + 3abc $
$ = {a^3} + {b^3} + {c^3} + 3ab\left( {a + c} \right) + 3ac\left( {a + c} \right) + 3{b^2}\left( {a + c} \right) + 3bc\left( {a + c} \right) $
$ = {a^3} + {b^3} + {c^3} + \left( {a + c} \right)\left( {3ab + 3ac + 3{b^2} + 3bc} \right) $
$ = {a^3} + {b^3} + {c^3} + \left( {a + c} \right)\left( {3a\left( {b + c} \right) + 3b\left( {b + c} \right)} \right) $
$ = {a^3} + {b^3} + {c^3} + \left( {a + c} \right)\left( {b + c} \right)\left( {3a + 3b} \right) $
$ = {a^3} + {b^3} + {c^3} + 3\left( {a + c} \right)\left( {b + c} \right)\left( {a + b} \right) $
Now, on transposing we get
$ {\left( {a + b + c} \right)^3}  {a^3}  {b^3}  {c^3} = 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) $
Introduction to NCERT Exemplar
NCERT Exemplar Class 9 Maths Solutions Chapter 2 'Polynomials' will give you data in regards to polynomials and their applications. The part of NCERT Exemplar Class 9 Maths incorporates various topics like entire numbers, integers, rational numbers, kinds of polynomials, zeros of polynomials, and so forth This part contains 4 activities and the evaluations of the around 60 model problems.
Polynomials of Class 9 NCERT comprise of very clear cut theory for students to do more practice that needs to solve questions given in the model. Chapter 2 comprises issues dependent on Factorization of Polynomials and Algebraic Identities and polynomials in one variable, Zeros of a Polynomial. Another significant formula you will comprehend is Remainder Theorem.
Importance of NCERT Class 9 Maths Exemplar Book
The NCERT Class 9 Maths Exemplar book comprises the important questions according to the assessment perspective. These NCERT Exemplar questions and answers for CBSE Class 9 Maths Chapter 2 are planned by the experts of maths who target at keeping each idea of the chapter clear for you. The experts at Vedantu have given you stepwise NCERT Exemplar arrangements so you can comprehend the ideas and score more marks in your CBSE tests.
FAQs on NCERT Exemplar for Class 9 Maths Chapter 2  Polynomials (Book Solutions)
1. What are the important tips for students to study in NCERT Class 9 Maths Chapter 2?
Problems in NCERT Exemplar Class 9 Maths Exemplar Chapter 2 will examine various concepts related to polynomials. In the word 'polynomials', 'poly' signifies 'many' and 'nomials' signifies 'term'. As far as mathematics is concerned, a polynomial contains factors that are referred to as coefficients and indeterminates as well. A coefficient includes various activities like addition, subtraction, no negative number type of factors just as multiplication. These are some important tips that Class 9 students should keep in mind while preparing for their Maths board examinations.
2. What do Polynomial and Constant Polynomials refer to in Class 9 Maths chapter 2 solutions?
The reference is as follows:
Polynomials :
Polynomials allude to the expressions that contain at least one term alongside a nonzero coefficient. Polynomials can have more than one term. In a polynomial, every single articulation is referred to as a term.
Constant Polynomials :
Whereas, At the point when the real numbers are showcased as polynomials like 367 are additionally polynomials that don't contain any factors, these are known as constant polynomials. Apart from this, a constant polynomial which refers to 0 is known as a zero polynomial.
3. What is Polynomial Function in NCERT Class 9 maths chapter 2?
A polynomial function is a capacity that includes just no negative numbers or just the positive numbers types of any factor in a situation like a quadratic condition, cubic condition, and so on. You can also consider a polynomial function as a polynomial articulation that is characterized by its degree. You can address a polynomial as p{x}. There are many polynomial functions that depend on the level of the polynomial and some of them are:
Zero polynomial function
Quadratic polynomial function
Linear polynomial function
4. How many questions are there in each exercise of NCERT Exemplar for Class 9 maths chapter 2?
The first exercise of NCERT Exemplar problems with answers for class 9 maths chapter 2 starts with the questions in which you need to distinguish if the polynomial is in 1 variable or not.
In the second exercise, you need to observe the values of the polynomials in various questions as these questions really look at your insightful abilities.
In activities 2.3.and 2.4, you will be checked assuming you can find, you will be able to distinguish in case a polynomial had the given factor.
Practice 2.5 will make you use the characters in the questions and if you will be able to factorize each question given.
5. Why should we use NCERT Exemplar Solutions Maths Chapter 2 – Polynomials By Vedantu?
Vedantu f urnishes you with the ideal NCERT Exemplar problems and answers for the class 9 maths chapter to guarantee that you can have the simplest access to the quality study material. We completely consider the significance of a subject and the topic that will be discussed. All the NCERT Exemplar problems with solutions guarantee that you can, without much of a stretch, revise the important points while you have your CBSE tests. Our team is sufficiently proficient to give you the best solutions free of cost!
NCERT Solutions Class 9 Maths Chapter 2 Polynomials
NCERT solutions for class 9 maths Chapter 2 Polynomials are all about the basics of polynomials like the different types of polynomials, finding roots, or solutions to a polynomial equation. Polynomials are algebraic expressions having one variable or more. These NCERT solutions class 9 maths Chapter 2 also explain the remainder theorem and factor theory of polynomials in detail, the algebraic identities, and polynomials of various degrees.
Class 9 Maths NCERT Solutions Chapter 2 polynomials illustrate the difference between linear, quadratic, and cubic polynomials. Important theorems mentioned are the Remainder theorem and the Factor theorem, which help identify the factors of a polynomial. Students can access the solutions from the pdf links given below and also find some of these in the exercises given below.
 NCERT Solutions Class 9 Maths Chapter 2 Ex 2.1
 NCERT Solutions Class 9 Maths Chapter 2 Ex 2.2
 NCERT Solutions Class 9 Maths Chapter 2 Ex 2.3
 NCERT Solutions Class 9 Maths Chapter 2 Ex 2.4
 NCERT Solutions Class 9 Maths Chapter 2 Ex 2.5
NCERT Solutions for Class 9 Maths Chapter 2 PDF
The exercises related to identifying the type of polynomial, finding the roots or solution of a polynomial equation, and finding factors of the polynomial are available for free pdf download using the four links provided below:
☛ Download Class 9 Maths NCERT Solutions Chapter 2 Polynomials
NCERT Class 9 Maths Chapter 2 Download PDF
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials
These fundamental properties and theorems of polynomials form the building blocks for higher mathematics. Thus, it is very important to master the fundamentals by solving many different example exercises using the links provided above. These NCERT Solution exercises will help understand the properties of polynomials better, as well as how to utilize them. Chapterwise detailed analysis of NCERT Solutions Class 9 Maths Chapter 2 Polynomials is given below.
 Class 9 Maths Chapter 2 Ex 2.1  21 Questions
 Class 9 Maths Chapter 2 Ex 2.2  22 Questions
 Class 9 Maths Chapter 2 Ex 2.3  7 Questions
 Class 9 Maths Chapter 2 Ex 2.4  7 Questions
 Class 9 Maths Chapter 2 Ex 2.5  41 Questions
☛ Download Class 9 Maths Chapter 2 NCERT Book
Topics Covered: The topics that are covered under the chapter on polynomials include an explanation of polynomials as a special set of algebraic equations, different types of polynomials, solutions of polynomial equations, factor theorem, and remainder theorem. Also, these class 9 maths NCERT solutions Chapter 2 define the algebraic identities , which help in factorizing the algebraic equations.
Total Questions: Class 9 Maths Chapter 2 Polynomials consists of a total of 45 questions, of which 31 are easy, 9 are moderate, and 5 are long answer type questions.
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List of Formulas in NCERT Solutions Class 9 Maths Chapter 2
NCERT solutions class 9 maths Chapter 2 covers lots of important concepts crucial for understanding higher grade maths. By learning to factorize a polynomial expression, one can find the roots of the polynomial equation. This is a relatively simple process that can greatly improve an individual's understanding of polynomial equations. Some important algebraic identities or formulas which help in factorization and are covered in NCERT solutions for class 9 maths chapter 2 are given below.
 ( x + y ) 2 = x 2 + 2xy + y 2
 ( x + y ) 3 = x 3 + y 3 + 3xy (x+y)
Important Questions for Class 9 Maths NCERT Solutions Chapter 2
CBSE Important Questions for Class 9 Maths Chapter 2 Exercise 2.1 

CBSE Important Questions for Class 9 Maths Chapter 2 Exercise 2.2 

CBSE Important Questions for Class 9 Maths Chapter 2 Exercise 2.3 

CBSE Important Questions for Class 9 Maths Chapter 2 Exercise 2.4 

CBSE Important Questions for Class 9 Maths Chapter 2 Exercise 2.5 

Video Solutions for Class 9 Maths NCERT Chapter 2
NCERT Video Solutions for Class 9 Maths Chapter 2  

Video Solutions for Class 9 Maths Exercise 2.1  
Video Solutions for Class 9 Maths Exercise 2.2  
Video Solutions for Class 9 Maths Exercise 2.3  
Video Solutions for Class 9 Maths Exercise 2.4  
Video Solutions for Class 9 Maths Exercise 2.5  
FAQs on NCERT Solutions Class 9 Maths Chapter 2
How cbse students can utilize ncert solutions class 9 maths chapter 2 effectively.
Algebra forms the basis of higher mathematical studies. Hence, students should focus on the important terms defined in this chapter, like the degree of a polynomial, the difference between constant and variable, to get a clear understanding of the polynomials. This will help them to make their base strong to appear for their board exams and face any kind of difficult questions.
Why are Class 9 Maths NCERT Solutions Chapter 2 Important?
The NCERT Solutions Class 9 Maths Chapter 2 includes a detailed explanation of the remainder and the factor theorem, which hold an important place in algebra. Also, the crucial algebraic identities are discussed in an elaborate manner with plenty of questions to solve for the students. A list of all key equations and concepts is available at the end of the chapter. This is a significant benefit because students can use this list whenever required instead of figuring it out from between the lengthy chapter text. Overall, these solutions cover all of the major concepts, approaches, and formulas, making them of utmost importance for class 9 math students.
How Many Questions are there in NCERT Solutions Class 9 Maths Chapter 2 Polynomials?
Overall the NCERT Solutions Class 9 Maths Chapter 2 has 98 questions that can be categorized as easy, medium, and difficult ones. Roughly 70 questions are straightforward and easy to solve, 20 questions are of medium difficulty level while 8 would require some thinking as they are longform questions.
What are the Important Topics Covered in NCERT Solutions Class 9 Maths Chapter 2?
The important topics that are covered under the NCERT Solutions Class 9 Maths Chapter 2 include the basic understanding of polynomials, the components of algebraic expressions, and their definitions. The chapter focuses on the types of polynomials and how to solve them, with special emphasis on factor and remainder theorem and the algebraic entities.
What are the Important Formulas in NCERT Solutions Class 9 Maths Chapter 2?
Since the NCERT Solutions Class 9 Maths Chapter 2 covers the polynomials from their basic structure, several definitions of important terms have been explained with their formulas, like the factor and the remainder theorem. But the most important formula would be the algebraic identities as they help in factorization itself. For example, (a + b) 2 = a 2 + 2ab + b 2
Do I Need to Practice all Questions Provided in NCERT Solutions Class 9 Maths Polynomials?
NCERT Solutions Class 9 Maths Polynomials encompass a variety of questions that explore all the algebraic concepts related to polynomials. Hence, it would be good if the students make use of this resource and start practicing by solving the examples first, which will help them in getting an idea of what steps are to be followed when questions related to polynomials are solved.
CBSE Expert
CBSE Case Study Questions Class 9 Maths Chapter 2 Polynomials PDF Download
CBSE Case Study Questions Class 9 Maths Chapter 2 Polynomials PDF Download are very important to solve for your exam. Class 9 Maths Chapter 2 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Case Study Questions Class 9 Maths Chapter 2 Polynomials
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Polynomials Case Study Questions With Answers
Case study questions class 9 maths chapter 2.
Case Study/PassageBased Questions
Case Study 1. Ankur and Ranjan start a new business together. The amount invested by both partners together is given by the polynomial p(x) = 4x 2 + 12x + 5, which is the product of their individual shares.
Coefficient of x 2 in the given polynomial is (a) 2 (b) 3 (c) 4 (d) 12
Answer: (c) 4
Total amount invested by both, if x = 1000 is (a) 301506 (b)370561 (c) 4012005 (d)490621
Answer: (c) 4012005
The shares of Ankur and Ranjan invested individually are (a) (2x + 1),(2x + 5)(b) (2x + 3),(x + 1) (c) (x + 1),(x + 3) (d) None of these
Answer: (a) (2x + 1),(2x + 5)
Name the polynomial of amounts invested by each partner. (a) Cubic (b) Quadratic (c) Linear (d) None of these
Answer: (c) Linear
Find the value of x, if the total amount invested is equal to 0. (a) –1/2 (b) –5/2 (c) Both (a) and (b) (d) None of these
Answer: (c) Both (a) and (b)
Case Study 2. One day, the principal of a particular school visited the classroom. The class teacher was teaching the concept of a polynomial to students. He was very much impressed by her way of teaching. To check, whether the students also understand the concept taught by her or not, he asked various questions to students. Some of them are given below. Answer them
Which one of the following is not a polynomial? (a) 4x 2 + 2x – 1 (b) y+3/y (c) x 3 – 1 (d) y 2 + 5y + 1
Answer: (b) y+3/y
The polynomial of the type ax 2 + bx + c, a = 0 is called (a) Linear polynomial (b) Quadratic polynomial (c) Cubic polynomial (d) Biquadratic polynomial
Answer: (a) Linear polynomial
The value of k, if (x – 1) is a factor of 4x 3 + 3x 2 – 4x + k, is (a) 1 (b) –2 (c) –3 (d) 3
Answer: (c) –3
If x + 2 is the factor of x 3 – 2ax 2 + 16, then value of a is (a) –7 (b) 1 (c) –1 (d) 7
Answer: (b) 1
The number of zeroes of the polynomial x 2 + 4x + 2 is (a) 1 (b) 2 (c) 3 (d) 4
Answer: (b) 2
Case Study 3. Amit and Rahul are friends who love collecting stamps. They decide to start a stamp collection club and contribute funds to purchase new stamps. They both invest a certain amount of money in the club. Let’s represent Amit’s investment by the polynomial A(x) = 3x^2 + 2x + 1 and Rahul’s investment by the polynomial R(x) = 2x^2 – 5x + 3. The sum of their investments is represented by the polynomial S(x), which is the sum of A(x) and R(x).
Q1. What is the coefficient of x^2 in Amit’s investment polynomial A(x)? (a) 3 (b) 2 (c) 1 (d) 0
Answer: (a) 3
Q2. What is the constant term in Rahul’s investment polynomial R(x)? (a) 2 (b) 5 (c) 3 (d) 0
Answer: (c) 6
Q3. What is the degree of the polynomial S(x), representing the sum of their investments? (a) 4 (b) 3 (c) 2 (d) 1
Answer: (c) 2
Q4. What is the coefficient of x in the polynomial S(x)? (a) 7 (b) 3 (c) 0 (d) 5
Answer: (b) 3
Q5. What is the sum of their investments, represented by the polynomial S(x)? (a) 5x^2 + 7x + 4 (b) 5x^2 – 3x + 4 (c) 5x^2 – 3x + 5 (d) 5x^2 + 7x + 5
Answer: (b) 5x^2 – 3x + 4
Case Study 4. A school is organizing a fundraising event to support a local charity. The students are divided into three groups: Group A, Group B, and Group C. Each group is responsible for collecting donations from different areas of the town.
Group A consists of 30 students and each student is expected to collect ‘x’ amount of money. The polynomial representing the total amount collected by Group A is given as A(x) = 2x^2 + 5x + 10.
Group B consists of 20 students and each student is expected to collect ‘y’ amount of money. The polynomial representing the total amount collected by Group B is given as B(y) = 3y^2 – 4y + 7.
Group C consists of 40 students and each student is expected to collect ‘z’ amount of money. The polynomial representing the total amount collected by Group C is given as C(z) = 4z^2 + 3z – 2.
Q1. What is the coefficient of x in the polynomial A(x)? (a) 2 (b) 5 (c) 10 (d) 0
Answer: (b) 5
Q2. What is the degree of the polynomial B(y)? (a) 2 (b) 3 (c) 4 (d) 1
Answer: (b) 3
Q3. What is the constant term in the polynomial C(z)? (a) 4 (b) 3 (c) 2 (d) 0
Answer: (c) 2
Q4. What is the sum of the coefficients of the polynomial A(x)? (a) 2 (b) 5 (c) 10 (d) 17
Answer: (c) 10
Q5. What is the total number of students in all three groups combined? (a) 30 (b) 20 (c) 40 (d) 90
Answer: (c) 40
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CBSE Important Questions for Class 9 Maths pdf
Important questions for class 9 maths chapter2 polynomials.
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Jul 21, 2022, 16:45 IST
Find below Important questions of class 9 maths of chapter Polynomials prepared by the academic team of Physics Wallah. All important questions of chapter Polynomials class 9 maths are uploaded in pdf form with detail stepbystep solutions.
Do solve NCERT questions and for reference use NCERT solutions for class 9 Maths prepared by Physics Wallah team. Students can also access the Class 9 Polynomials notes from here. For additional information related to the subject you can check the Maths Formula section and If any students need to take the online test to check their concepts or undertstanding then they can visit Quiz for Polynomials .
Download free pdf for CBSE Important Questions for Class 9 Maths Chapter Polynomials
Related link.
 chapter1 Number Systems
 Chapter2 Polynomials
 Chapter3 Coordinate Geometry
 Chapter4 Linear Equations in Two Variables
 chapter5 Introduction to Euclids Geometry
 chapter6 Lines and Angles
 chapter7 Triangles
 Chapter8 Quadrilaterals
 Chapter9 Areas of Parallelograms and Triangles
 Chapter10 Circles
 Chapter13 Surface Areas and Volumes
 Chapter14 Statistics
 Chapter15 Probability
 Chapter12 Heron’s Formula
 chapter11 Constructions
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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials
Ncert solutions for class 9 maths chapter 2 polynomials pdf download.
 Exercise 2.1 Chapter 2 Class 9 Maths NCERT Solutions
 Exercise 2.2 Chapter 2 Class 9 Maths NCERT Solutions
 Exercise 2.3 Chapter 2 Class 9 Maths NCERT Solutions
 Exercise 2.4 Chapter 2 Class 9 Maths NCERT Solutions
 Exercise 2.5 Chapter 2 Class 9 Maths NCERT Solutions
NCERT Solutions for Class 9 Maths Chapters:
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NCERT Solutions for Class 9 Maths Chapter 2 Free PDF Download
Ncert solutions for class 9 maths chapter 2 polynomials.
This article is about NCERT Solutions for Class 9 Maths Chapter 2. NCERT Solutions gives you a great understanding of the topic polynomials. It helps the students by giving them a detailed explanation. So it makes them clear about the topic. Polynomial is a topic which confuses students a lot. However, NCERT Solutions gives the students a better understanding. It helps them score better marks in their class 9 exams.
Since Class 9 is the stepping stone for board examinations it makes it another crucial year for the students.NCERT Solutions for Class 9 Maths Chapter 2 Polynomials helps you in your time of need. It gives you access solutions in a few seconds. The solutions are one click away.
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CBSE Class 10 Mathematics Chapter 2 Polynomials NCERT Solutions
This chapter teaches about a particular type of algebraic expression called polynomials. Moreover, the students will get to know the basic terminology. In addition to difficult topics like Remainder theorem and factor, the theorem is explained. The explanation is in the simplest way. Therefore this makes our solutions stand out of the crowd.
Subtopics covered under NCERT Solutions for Class 9 Maths Chapter 2
2.1 introduction.
NCERT Solutions for class 9 chapter 2 polynomials gives you a thorough understanding of the topic. Moreover, our experts give proper attention to the Remainder Theorem and Factor Theorem uses and methods. These subheadings allow the student to grasp the topics in a much easier way. Our teachers explain polynomial terminology and different algebraic identities.
2.2 Polynomials in one variable
In this section, we study the polynomials in one variable. Polynomials in one variable contain only one type of variable. They are generally x, y, z. You will get to know more about our solutions for class 9 chapter 2 polynomials.
2.3 Zeroes of a polynomials
Zeroes of a polynomial are the answer when we assign a particular value to a variable.
2.4 Remainder Theorem
Remainder Theorem helps in the division of two polynomials. This section gives you a detailed explanation of the method. It also contains an ample amount of examples. These examples will help you understand the method.
2.5 Factorization of polynomials
The factorization of the polynomial takes place with factor theorem. The factor theorem takes out the factor of a polynomial. Examples explain the Method of factorization in a detailed format.
2.6 Algebraic Identities
Here we study the various algebraic identities and their uses. An algebraic identity is an equation which holds true for all the values of the variables. The solution explains different identities in the chapter. There are a total of 8 identities. Separate examples give a good explanation of the identities.
2.7 Summary
The summary gives a brief description of all the topics of the chapter. It gives a quick glance to the student at the time of revision. It helps the student to understand the essential meaning of the different topics.
You can download NCERT Solutions for Class 9 Maths Chapter 2 PDF Download for free here
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CBSE Class 9th Maths 2023 : 30 Most Important Case Study Questions with Answers; Download PDF
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CBSE Class 9 Maths exam 202223 will have a set of questions based on case studies in the form of MCQs. CBSE Class 9 Maths Question Bank on Case Studies given in this article can be very helpful in understanding the new format of questions.
Each question has five subquestions, each followed by four options and one correct answer. Students can easily download these questions in PDF format and refer to them for exam preparation.
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Case Study Questions for Class 9 Maths Chapter 2 Polynomials
 Last modified on: 2 years ago
 Reading Time: 3 Minutes
Case Study Questions
Question 1:
On one day, principal of a particular school visited the classroom. Class teacher was teaching the concept of polynomial to students. He was very much impressed by her way of teaching. To check, whether the students also understand the concept taught by her or not, he asked various questions to students. Some of them are given below. Answer them.
(i) Which one of the following is not a polynomial? (a) 4x 2 + 2x – 1 (b) y+ (3/y) (c) x 3 – 1 (d) y 2 + 5y + 1
(ii) The polynomial of the type ax 2 + bx + c, a = 0 is called (a) Linear polynomial (b) Quadratic polynomial (c) Cubic polynomial (d) Biquadratic polynomial
(iii) The value of k, if (x – 1) is a factor of 4x 3 + 3x 2 – 4x + k, is (a) 1 (b) –2 (c) –3 (d) 3
(iv) If x + 2 is the factor of x 3 – 2ax 2 + 16, then value of a is (a) –7 (b) 1 (c) –1 (d) 7
(v) The number of zeroes of the polynomial x 2 + 4x + 2 is (a) 1 (b) 2 (c) 3 (d) 4
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 CBSE Class 9 Mathematics...
CBSE Class 9 Mathematics Case Study Questions
Table of Contents
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Significance of Mathematics in Class 9
Mathematics is an important subject for students of all ages. It helps students to develop problemsolving and criticalthinking skills, and to think logically and creatively. In addition, mathematics is essential for understanding and using many other subjects, such as science, engineering, and finance.
CBSE Class 9 is an important year for students, as it is the foundation year for the Class 10 board exams. In Class 9, students learn many important concepts in mathematics that will help them to succeed in their board exams and in their future studies. Therefore, it is essential for students to understand and master the concepts taught in Class 9 Mathematics .
Case studies in Class 9 Mathematics
A case study in mathematics is a detailed analysis of a particular mathematical problem or situation. Case studies are often used to examine the relationship between theory and practice, and to explore the connections between different areas of mathematics. Often, a case study will focus on a single problem or situation and will use a variety of methods to examine it. These methods may include algebraic, geometric, and/or statistical analysis.
Example of Case study questions in Class 9 Mathematics
The Central Board of Secondary Education (CBSE) has included case study questions in the Class 9 Mathematics paper. This means that Class 9 Mathematics students will have to solve questions based on reallife scenarios. This is a departure from the usual theoretical questions that are asked in Class 9 Mathematics exams.
The following are some examples of case study questions from Class 9 Mathematics:
Class 9 Mathematics Case study question 1
There is a square park ABCD in the middle of Saket colony in Delhi. Four children Deepak, Ashok, Arjun and Deepa went to play with their balls. The colour of the ball of Ashok, Deepak, Arjun and Deepa are red, blue, yellow and green respectively. All four children roll their ball from centre point O in the direction of XOY, X’OY, X’OY’ and XOY’ . Their balls stopped as shown in the above image.
Answer the following questions:
Answer Key:
Class 9 Mathematics Case study question 2
 Now he told Raju to draw another line CD as in the figure
 The teacher told Ajay to mark ∠ AOD as 2z
 Suraj was told to mark ∠ AOC as 4y
 Clive Made and angle ∠ COE = 60°
 Peter marked ∠ BOE and ∠ BOD as y and x respectively
Now answer the following questions:
 2y + z = 90°
 2y + z = 180°
 4y + 2z = 120°
 (a) 2y + z = 90°
Class 9 Mathematics Case study question 3
 (a) 31.6 m²
 (c) 513.3 m³
 (b) 422.4 m²
Class 9 Mathematics Case study question 4
How to Answer Class 9 Mathematics Case study questions
To crack case study questions, Class 9 Mathematics students need to apply their mathematical knowledge to reallife situations. They should first read the question carefully and identify the key information. They should then identify the relevant mathematical concepts that can be applied to solve the question. Once they have done this, they can start solving the Class 9 Mathematics case study question.
Students need to be careful while solving the Class 9 Mathematics case study questions. They should not make any assumptions and should always check their answers. If they are stuck on a question, they should take a break and come back to it later. With some practice, the Class 9 Mathematics students will be able to crack case study questions with ease.
Class 9 Mathematics Curriculum at Glance
At the secondary level, the curriculum focuses on improving students’ ability to use Mathematics to solve realworld problems and to study the subject as a separate discipline. Students are expected to learn how to solve issues using algebraic approaches and how to apply their understanding of simple trigonometry to height and distance problems. Experimenting with numbers and geometric forms, making hypotheses, and validating them with more observations are all part of Math learning at this level.
The suggested curriculum covers number systems, algebra, geometry, trigonometry, mensuration, statistics, graphing, and coordinate geometry, among other topics. Math should be taught through activities that include the use of concrete materials, models, patterns, charts, photographs, posters, and other visual aids.
CBSE Class 9 Mathematics (Code No. 041)
I  NUMBER SYSTEMS  10 
II  ALGEBRA  20 
III  COORDINATE GEOMETRY  04 
IV  GEOMETRY  27 
V  MENSURATION  13 
VI  STATISTICS & PROBABILITY  06 
Class 9 Mathematics question paper design
The CBSE Class 9 mathematics question paper design is intended to measure students’ grasp of the subject’s fundamental ideas. The paper will put their problemsolving and analytical skills to the test. Class 9 mathematics students are advised to go through the question paper pattern thoroughly before they start preparing for their examinations. This will help them understand the paper better and enable them to score maximum marks. Refer to the given Class 9 Mathematics question paper design.
QUESTION PAPER DESIGN (CLASS 9 MATHEMATICS)
1.  Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers. Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas  43  54 
2.  Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.  19  24 
3.  Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations Present and defend opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria. Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions  18  22 
80  100 
myCBSEguide: Blessing in disguise
Class 9 is an important milestone in a student’s life. It is the last year of high school and the last chance to score well in the CBSE board exams. myCBSEguide is the perfect platform for students to get started on their preparations for Class 9 Mathematics. myCBSEguide provides comprehensive study material for all subjects, including practice questions, sample papers, case study questions and mock tests. It also offers tips and tricks on how to score well in exams. myCBSEguide is the perfect door to enter for class 9 CBSE preparations.
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16 thoughts on “CBSE Class 9 Mathematics Case Study Questions”
This method is not easy for me
aarti and rashika are two classmates. due to exams approaching in some days both decided to study together. during revision hour both find difficulties and they solved each other’s problems. aarti explains simplification of 2+ ?2 by rationalising the denominator and rashika explains 4+ ?2 simplification of (v10?5)(v10+ ?5) by using the identity (a – b)(a+b). based on above information, answer the following questions: 1) what is the rationalising factor of the denominator of 2+ ?2 a) 2?2 b) 2?2 c) 2+ ?2 by rationalising the denominator of aarti got the answer d) a) 4+3?2 b) 3+?2 c) 3?2 4+ ?2 2+ ?2 d) 2?3 the identity applied to solve (?10?5) (v10+ ?5) is a) (a+b)(a – b) = (a – b)² c) (a – b)(a+b) = a² – b² d) (ab)(a+b)=2(a² + b²) ii) b) (a+b)(a – b) = (a + b
MATHS PAAGAL HAI
All questions was easy but search ? hard questions. These questions was not comparable with cbse. It was totally wastage of time.
Where is search ? bar
maths is love
Can I have more questions without downloading the app.
I love math
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I am Riddhi Shrivastava… These questions was very good.. That’s it.. ..
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CBSE Case Study Questions for Class 9 Maths  Pdf PDF Download
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CBSE Case Study Questions for Class 9 Maths
CBSE Case Study Questions for Class 9 Maths are a type of assessment where students are given a realworld scenario or situation and they need to apply mathematical concepts to solve the problem. These types of questions help students to develop their problemsolving skills and apply their knowledge of mathematics to reallife situations.
Chapter Wise Case Based Questions for Class 9 Maths
The CBSE Class 9 Case Based Questions can be accessed from Chapetrwise Links provided below:
Chapterwise casebased questions for Class 9 Maths are a set of questions based on specific chapters or topics covered in the maths textbook. These questions are designed to help students apply their understanding of mathematical concepts to realworld situations and events.
Chapter 1: Number System
 Case Based Questions: Number System
 Chapter 2: Polynomial
 Case Based Questions: Polynomial
Chapter 3: Coordinate Geometry
 Case Based Questions: Coordinate Geometry
Chapter 4: Linear Equations
 Case Based Questions: Linear Equations  1
 Case Based Questions: Linear Equations 2
Chapter 5: Introduction to Euclid’s Geometry
 Case Based Questions: Lines and Angles
Chapter 7: Triangles
 Case Based Questions: Triangles
Chapter 8: Quadrilaterals
 Case Based Questions: Quadrilaterals  1
 Case Based Questions: Quadrilaterals  2
Chapter 9: Areas of Parallelograms
 Case Based Questions: Circles
Chapter 11: Constructions
 Case Based Questions: Constructions
Chapter 12: Heron’s Formula
 Case Based Questions: Heron’s Formula
Chapter 13: Surface Areas and Volumes
 Case Based Questions: Surface Areas and Volumes
Chapter 14: Statistics
 Case Based Questions: Statistics
Chapter 15: Probability
 Case Based Questions: Probability
Weightage of Case Based Questions in Class 9 Maths
Why are Case Study Questions important in Maths Class 9?
 Enhance critical thinking: Case study questions require students to analyze a reallife scenario and think critically to identify the problem and come up with possible solutions. This enhances their critical thinking and problemsolving skills.
 Apply theoretical concepts: Case study questions allow students to apply theoretical concepts that they have learned in the classroom to reallife situations. This helps them to understand the practical application of the concepts and reinforces their learning.
 Develop decisionmaking skills: Case study questions challenge students to make decisions based on the information provided in the scenario. This helps them to develop their decisionmaking skills and learn how to make informed decisions.
 Improve communication skills: Case study questions often require students to present their findings and recommendations in written or oral form. This helps them to improve their communication skills and learn how to present their ideas effectively.
 Enhance teamwork skills: Case study questions can also be done in groups, which helps students to develop teamwork skills and learn how to work collaboratively to solve problems.
In summary, case study questions are important in Class 9 because they enhance critical thinking, apply theoretical concepts, develop decisionmaking skills, improve communication skills, and enhance teamwork skills. They provide a practical and engaging way for students to learn and apply their knowledge and skills to reallife situations.
Class 9 Maths Curriculum at Glance
The Class 9 Maths curriculum in India covers a wide range of topics and concepts. Here is a brief overview of the Maths curriculum at a glance:
 Number Systems: Students learn about the real number system, irrational numbers, rational numbers, decimal representation of rational numbers, and their properties.
 Algebra: The Algebra section includes topics such as polynomials, linear equations in two variables, quadratic equations, and their solutions.
 Coordinate Geometry: Students learn about the coordinate plane, distance formula, section formula, and slope of a line.
 Geometry: This section includes topics such as Euclid’s geometry, lines and angles, triangles, and circles.
 Trigonometry: Students learn about trigonometric ratios, trigonometric identities, and their applications.
 Mensuration: This section includes topics such as area, volume, surface area, and their applications.
 Statistics and Probability: Students learn about measures of central tendency, graphical representation of data, and probability.
The Class 9 Maths curriculum is designed to provide a strong foundation in mathematics and prepare students for higher education in the field. The curriculum is structured to develop critical thinking, problemsolving, and analytical skills, and to promote the application of mathematical concepts in reallife situations. The curriculum is also designed to help students prepare for competitive exams and develop a strong mathematical base for future academic and professional pursuits.
Students can also access Case Based Questions of all subjects of CBSE Class 9
 Case Based Questions for Class 9 Science
 Case Based Questions for Class 9 Social Science
 Case Based Questions for Class 9 English
 Case Based Questions for Class 9 Hindi
 Case Based Questions for Class 9 Sanskrit
Frequently Asked Questions (FAQs) on Case Based Questions for Class 9 Maths
What is casebased questions.
CaseBased Questions (CBQs) are openended problem solving tasks that require students to draw upon their knowledge of Maths concepts and processes to solve a novel problem. CBQs are often used as formative or summative assessments, as they can provide insights into how students reason through and apply mathematical principles in realworld problems.
What are casebased questions in Maths?
Casebased questions in Maths are problemsolving tasks that require students to apply their mathematical knowledge and skills to realworld situations or scenarios.
What are some common types of casebased questions in class 9 Maths?
Common types of casebased questions in class 9 Maths include word problems, realworld scenarios, and mathematical modeling tasks.
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FAQs on CBSE Case Study Questions for Class 9 Maths  Pdf
1. What are case study questions in CBSE Class 9 Maths? 
2. How are case study questions different from regular math questions in Class 9? 
3. Why are case study questions important in Class 9 Maths? 
4. How much weightage do case study questions have in the Class 9 Maths exam? 
5. Can you provide some tips to effectively answer case study questions in Class 9 Maths? 
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CBSE Class 9 Maths Most Important Case Study Based Questions With Solution
Cbse class 9 mathematics case study questions.
In this post I have provided CBSE Class 9 Maths Case Study Based Questions With Solution. These questions are very important for those students who are preparing for their final class 9 maths exam.
All these questions provided in this article are with solution which will help students for solving the problems. Dear students need to practice all these questions carefully with the help of given solutions.
As you know CBSE Class 9 Maths exam will have a set of cased study based questions in the form of MCQs. CBSE Class 9 Maths Question Bank given in this article can be very helpful in understanding the new format of questions for new session.
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Case studies in class 9 mathematics.
The Central Board of Secondary Education (CBSE) has included case study based questions in the Class 9 Mathematics paper in current session. According to new pattern CBSE Class 9 Mathematics students will have to solve case based questions. This is a departure from the usual theoretical conceptual questions that are asked in Class 9 Maths exam in this year.
Each question provided in this post has five subquestions, each followed by four options and one correct answer. All CBSE Class 9th Maths Students can easily download these questions in PDF form with the help of given download Links and refer for exam preparation.
There is many more free study materials are available at Maths And Physics With Pandey Sir website. For many more books and free study material all of you can visit at this website.
Given Below Are CBSE Class 9th Maths Case Based Questions With Their Respective Download Links.
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Case Study Questions for Class 9 Maths
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Are you preparing for your Class 9 Maths board exams and looking for an effective study resource? Well, you’re in luck! In this article, we will provide you with a collection of Case Study Questions for Class 9 Maths specifically designed to help you excel in your exams. These questions are carefully curated to cover various mathematical concepts and problemsolving techniques. So, let’s dive in and explore these valuable resources that will enhance your preparation and boost your confidence.
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CBSE Class 9 Maths Board Exam will have a set of questions based on case studies in the form of MCQs. The CBSE Class 9 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.
If you want to want to prepare all the tough, tricky & difficult questions for your upcoming exams, this is where you should hang out. CBSE Case Study Questions for Class 9 will provide you with detailed, latest, comprehensive & confidenceinspiring solutions to the maximum number of Case Study Questions covering all the topics from your NCERT Text Books !
CBSE Class 9th – MATHS: Chapterwise Case Study Question & Solution
Case study questions are a form of examination where students are presented with reallife scenarios that require the application of mathematical concepts to arrive at a solution. These questions are designed to assess students’ problemsolving abilities, critical thinking skills, and understanding of mathematical concepts in practical contexts.
Chapterwise Case Study Questions for Class 9 Maths
Case study questions play a crucial role in the field of mathematics education. They provide students with an opportunity to apply theoretical knowledge to realworld situations, thereby enhancing their comprehension of mathematical concepts. By engaging with case study questions, students develop the ability to analyze complex problems, make connections between different mathematical concepts, and formulate effective problemsolving strategies.
 Case Study Questions for Chapter 1 Number System
 Case Study Questions for Chapter 2 Polynomials
 Case Study Questions for Chapter 3 Coordinate Geometry
 Case Study Questions for Chapter 4 Linear Equations in Two Variables
 Case Study Questions for Chapter 5 Introduction to Euclid’s Geometry
 Case Study Questions for Chapter 6 Lines and Angles
 Case Study Questions for Chapter 7 Triangles
 Case Study Questions for Chapter 8 Quadilaterals
 Case Study Questions for Chapter 9 Areas of Parallelograms and Triangles
 Case Study Questions for Chapter 10 Circles
 Case Study Questions for Chapter 11 Constructions
 Case Study Questions for Chapter 12 Heron’s Formula
 Case Study Questions for Chapter 13 Surface Area and Volumes
 Case Study Questions for Chapter 14 Statistics
 Case Study Questions for Chapter 15 Probability
The above Case studies for Class 9 Mathematics will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Maths Case Studies have been developed by experienced teachers of schools.studyrate.in for benefit of Class 10 students.
 Class 9 Science Case Study Questions
 Class 9 Social Science Case Study Questions
How to Approach Case Study Questions
When tackling case study questions, it is essential to adopt a systematic approach. Here are some steps to help you approach and solve these types of questions effectively:
 Read the case study carefully: Understand the given scenario and identify the key information.
 Identify the mathematical concepts involved: Determine the relevant mathematical concepts and formulas applicable to the problem.
 Formulate a plan: Devise a plan or strategy to solve the problem based on the given information and mathematical concepts.
 Solve the problem step by step: Apply the chosen approach and perform calculations or manipulations to arrive at the solution.
 Verify and interpret the results: Ensure the solution aligns with the initial problem and interpret the findings in the context of the case study.
Tips for Solving Case Study Questions
Here are some valuable tips to help you effectively solve case study questions:
 Read the question thoroughly and underline or highlight important information.
 Break down the problem into smaller, manageable parts.
 Visualize the problem using diagrams or charts if applicable.
 Use appropriate mathematical formulas and concepts to solve the problem.
 Show all the steps of your calculations to ensure clarity.
 Check your final answer and review the solution for accuracy and relevance to the case study.
Benefits of Practicing Case Study Questions
Practicing case study questions offers several benefits that can significantly contribute to your mathematical proficiency:
 Enhances critical thinking skills
 Improves problemsolving abilities
 Deepens understanding of mathematical concepts
 Develops analytical reasoning
 Prepares you for reallife applications of mathematics
 Boosts confidence in approaching complex mathematical problems
Case study questions offer a unique opportunity to apply mathematical knowledge in practical scenarios. By practicing these questions, you can enhance your problemsolving abilities, develop a deeper understanding of mathematical concepts, and boost your confidence for the Class 9 Maths board exams. Remember to approach each question systematically, apply the relevant concepts, and review your solutions for accuracy. Access the PDF resource provided to access a wealth of case study questions and further elevate your preparation.
Q1: Can case study questions help me score better in my Class 9 Maths exams?
Yes, practicing case study questions can significantly improve your problemsolving skills and boost your performance in exams. These questions offer a practical approach to understanding mathematical concepts and their reallife applications.
Q2: Are the case study questions in the PDF resource relevant to the Class 9 Maths syllabus?
Absolutely! The PDF resource contains case study questions that align with the Class 9 Maths syllabus. They cover various topics and concepts included in the curriculum, ensuring comprehensive preparation.
Q3: Are the solutions provided for the case study questions in the PDF resource?
Yes, the PDF resource includes solutions for each case study question. You can refer to these solutions to validate your answers and gain a better understanding of the problemsolving process.
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Important Questions Class 9 Maths Chapter 2
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Important Questions Class 9 Mathematics Chapter 2 Polynomials.
Mathematics Chapter 2 of Class 9 is about Polynomials. Polynomial consists of two terms, namely Poly (meaning “many”) and Nominal (meaning “terms.”). A polynomial is explained as an expression which is composed of variables, constants and exponents that are combined using mathematical operations like addition, subtraction, multiplication and division (No division operation by a variable). Based on the number of terms present in the expression, it is classified as monomial, binomial, and trinomial.
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Our Mathematics experts believe that students must practice questions regularly to perform better in exams. For this purpose, they have prepared the Important Questions Class 9 Mathematics Chapter 2 to help students get access to questions from all the topics of the Polynomials. The questions are followed by their stepbystep answers, which will further help students to revise the chapter. The questions are curated from various sources such as the NCERT textbook and exemplar book, CBSE past years’ question papers, and other reference materials.
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Important Questions Class 9 Mathematics Chapter 2 – With Solutions
Our inhouse Mathematics faculty experts have collated a complete list of Important Questions Class 9 Mathematics Chapter 2 by referring to various sources. The subject experts have meticulously prepared illustration for individual questions that will enable students to comprehend the notions used in each question. Furthermore, the questions are selected in a way that would cover all the topics. So by practising from our question bank, students will be able to revise the chapter and understand their strong and weak topics. And enhance their preparation by further concentrating on weaker sections of the chapter.
Given below are a few of the questions and answers from our question bank of Important Questions Class 9 Mathematics Chapter 2:
Question 1: Calculate the value of 9x² + 4y² if xy = 6 and 3x + 2y = 12.
Answer 1: Consider the equation 3x + 2y = 12
Now, square both sides:
(3x + 2y)² = 12²
=> 9x² + 12xy + 4y² = 144
=>9x² + 4y² = 144 – 12xy
From the questions, xy = 6
9x² + 4y² = 144 – 72
Thus, the value of 9x² + 4y² = 72
Question 2:Evaluate the following using suitable identity
Answer 2: We can write 102 as 100+2
Using identity,(x+y) ³ = x ³ +y ³ +3xy(x+y)
(100+2) ³ =(100) ³ +2 ³ +(3×100×2)(100+2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
Question 3:Without any actual division, prove that the following 2x⁴
– 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.
[Hint: Factorise x² – 3x + 2]
Answer 3: x²3x+2
x(x2)1(x2)
Therefore,(x2)(x1)are the factors.
Considering (x2),
Then, p(x) becomes,
p(x)=2x⁴5x³+2x²x+2
p(2)=2(2)⁴5(2)³+2(2)²2+2
Therefore, (x2) is a factor.
Considering (x1),
p(1)=2(1)⁴5(1)³+2(1)²1+2
Therefore, (x1) is a factor.
Question 4: Using the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case
(i) p(x) = 2x³+x²–2x–1, g(x) = x+1
Answer 4 :p(x) = 2x³+x²–2x–1, g(x) = x+1
∴Zero of g(x) is 1.
p(−1) = 2(−1)³+(−1)²–2(−1)–1
∴By the given factor theorem, g(x) is a factor of p(x).
Question 5: Obtain an example of a monomial and a binomial having degrees of 82 and 99, respectively.
Answer 5: An example of a monomial having a degree of 82 = x⁸²
An example of a binomial having a required degree of 99 = x⁹⁹ + 7
Question 6: If the two x – 2 and x – ½ are the given factors of px ²
+ 5 x + r , show that p = r .
Answer 6: Given, f(x) = px²+5x+r and factors are x2, x – ½
Substituting x = 2 in place of the equation, we get
f(x) = px²+5x+r
f(2) = p(2)²+5(2)+r=0
= 4p + 10 + r = 0 … eq.(i)
Substituting x = ½ in place of the equation, we get,
f( ½ ) = p( ½ )² + 5( ½ ) + r =0
= p/4 + 5/2 + r = 0
= p + 10 + 4r = 0 … eq(ii)
On solving eq(i) and eq(ii),
4p + r = – 10 and p + 4r = – 10
the RHS of both equations are the same,
4p + r = p + 4r
Hence Proved.
Question 7: Identify constant, linear, quadratic, cubic and quartic polynomials from the following.
(i) – 7 + x
(iii) – ? ³
(iv) 1 – y – ? ³
(v) x – ? ³ + ?⁴
(vi) 1 + x + ?²
Answer 7: (i) – 7 + x
The degree of – 7 + x is 1.
Hence, it is a linear polynomial.
The degree of 6y is 1.
Therefore, it is a linear polynomial.
We know that the degree of – ? ³ is 3.
Therefore, it is a cubic polynomial.
We know that the degree of 1 – y – ? ³ is 3.
We know that the degree of x – ? ³ + ?⁴ is 4.
Therefore, it is a quartic polynomial.
We know that the degree of 1 + x + ?² is 2.
Therefore, it is a quadratic polynomial.
We know that the degree of 6?² is 2.
We know that 13 is a constant.
Therefore, it is a constant polynomial.
We know that the degree of –p is 1.
Question 8: Observe the value of the polynomial 5x – 4x² + 3 at x = 2 and x = –1.
Answer 8 : Let the polynomial be f(x) = 5x – 4x² + 3
Now, for x = 2,
f(2) = 5(2) – 4(2)² + 3
=> f(2) = 10 – 16 + 3 = –3
Or, the value of the polynomial 5x – 4x² + 3 at x = 2 is 3.
Similarly, for x = –1,
f(–1) = 5(–1) – 4(–1)² + 3
=> f(–1) = –5 –4 + 3 = 6
The value of the polynomial 5x – 4x² + 3 at x = 1 is 6.
Question 9:Expanding each of the following, using all the suitable identities:
(i) (x+2y+4z)²
(ii) (2x−y+z)²
(iii) (−2x+3y+2z)²
(iv) (3a –7b–c)²
(v) (–2x+5y–3z)²
Answer 9: (i) (x+2y+4z)²
Using identity, (x+y+z)² = x²+²+z²+2xy+2yz+2zx
Here, x = x
(x+2y+4z)² = x²+(2y)²+(4z)²+(2×x×2y)+(2×2y×4z)+(2×4z×x)
= x²+4y²+16z²+4xy+16yz+8xz
Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx
Here, x = 2x
(2x−y+z)² = (2x)²+(−y)²+z²+(2×2x×−y)+(2×−y×z)+(2×z×2x)
= 4x²+y²+z²–4xy–2yz+4xz
Here, x = −2x
(−2x+3y+2z)² = (−2x)²+(3y)²+(2z)²+(2×−2x×3y)+(2×3y×2z)+(2×2z×−2x)
= 4x²+9y²+4z²–12xy+12yz–8xz
Using identity (x+y+z)²= x²+y²+z²+2xy+2yz+2zx
Here, x = 3a
(3a –7b– c)² = (3a)²+(– 7b)²+(– c)²+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)
= 9a² + 49b² + c²– 42ab+14bc–6ca
Here, x = –2x
(–2x+5y–3z)² = (–2x)² + (5y)² + (–3z)² + (2 × –2x × 5y) + (2 × 5y× – 3z)+(2×–3z ×–2x)
= 4x²+25y² +9z²– 20xy–30yz+12zx
Question 10: If the polynomials az³ + 4z² + 3z – 4 and z³ – 4z + leave the same remainder when divided by z – 3, find the value of a.
Answer 10: Zero of the polynomial,
Hence, zero of g(z) = – 2a
Let p(z) = az³+4z²+3z4
Now, substituting the given value of z = 3 in p(z), we get,
p(3) = a (3)³ + 4 (3)² + 3 (3) – 4
⇒p(3) = 27a+36+94
⇒p(3) = 27a+41
Let h(z) = z³4z+a
Now, by substituting the value of z = 3 in h(z), we get,
h(3) = (3)³4(3)+a
⇒h(3) = 2712+a
⇒h(3) = 15+a
As per the question,
The two polynomials, p(z) and h(z), leave the same remainder when divided by z3
So, h(3)=p(3)
⇒15+a = 27a+41
⇒1541 = 27a – a
Question 11: Compute the perimeter of a rectangle whose area is 25x² – 35x + 12.
Answer 11: A rea of rectangle = 25x² – 35x + 12
We know the area of a rectangle = length × breadth
So, by factoring 25x² – 35x + 12, the length and breadth can be obtained.
25x² – 35x + 12 = 25x² – 15x – 20x + 12
=> 25x² – 35x + 12 = 5x(5x – 3) – 4(5x – 3)
=> 25x² – 35x + 12 = (5x – 3)(5x – 4)
Thus, the length and breadth of a rectangle are (5x – 3)(5x – 4).
So, the perimeter = 2(length + breadth)
Therefore, the perimeter of the given rectangle = 2[(5x – 3)+(5x – 4)]
= 2(5x – 3 + 5x – 4)
= 2(10x – 7)
= 20x – 14
Hence, the perimeter of the rectangle = 20x – 14
Question 12: 2x²+y²+²–2√2xy+4√2yz–8xz
Answer 12: Using identity, (x +y+z)² = x²+y²+z²+2xy+2yz+2zx
We can say that, x²+²+²+2xy+2yz+2zx = (x+y+z)²
2x²+y²+8z²–2√2xy+4√2yz–8xz
= (√2x)²+(y)²+(2√2z)²+(2×√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)
= (−√2x+y+2√2z)²
= (−√2x+y+2√2z)(−√2x+y+2√2z)
Question 13: If ? + 2? is a factor of ? ⁵ – 4?²?³ + 2? + 2? + 3, find a.
Answer 13: According to the question,
Let p(x) = x ⁵ – 4a²x³ + 2x + 2a + 3 and g(x) = x + 2a
⟹ x + 2a = 0
Hence, zero of g(x) = – 2a
As per the factor theorem,
If g(x) is a factor of p(x), then p( – 2a) = 0
So, substituting the value of x in p(x), we get,
p ( – 2a) = ( – 2a) ⁵ – 4a²( – 2a)³ + 2( – 2a) + 2a + 3 = 0
⟹ – 32a ⁵ + 32a ⁵ – 2a + 3 = 0
⟹ – 2a = – 3
Question 14: Find the value of x³+ y ³ + z ³ – 3xyz if x² + y² + z² = 83 and x + y + z = 1
Answer 14: Consider the equation x + y + z = 15
From algebraic identities, we know that (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(x + y + z)² = x² + y² + z² + 2(xy + yz + xz)
From the question, x² + y² + z²= 83 and x + y + z = 15
152 = 83 + 2(xy + yz + xz)
=> 225 – 83 = 2(xy + yz + xz)
Or, xy + yz + xz = 142/2 = 71
Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca),
x ³ + y ³ + z ³ – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz))
x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71
So, x ³ + y ³ + z ³ – 3xyz = 15(83 – 71)
=> x ³ + y ³ + z ³ – 3xyz = 15 × 12
Or, x ³ + y ³ + z ³ – 3xyz = 180
Question 15:Verify that:
(i) x³+y³ = (x+y)(x²–xy+y²)
(ii) x³–y³ = (x–y)(x²+xy+y²)
Answer 15: (i) x³+y³ = (x+y)(x²–xy+y²)
We know that (x+y)³= x³+y³+3xy(x+y)
⇒ x³+y³ = (x+y)³–3xy(x+y)
⇒ x³+y³ = (x+y)[(x+y)²–3xy]
Taking (x+y) common ⇒ x³+y³ = (x+y)[(x²+y²+2xy)–3xy]
⇒ x³+y³ = (x+y)(x²+y²–xy)
(ii) x³–y³ = (x–y)(x²+xy+y²)
We know that (x–y)³ = x³–y³–3xy(x–y)
⇒ x³−y³ = (x–y)³+3xy(x–y)
⇒ x³−y³ = (x–y)[(x–y)²+3xy]
Taking (x+y) common ⇒ x³−y³ = (x–y)[(x²+y²–2xy)+3xy]
⇒ x³+y³ = (x–y)(x²+y²+xy)
Question 16: For what value of m is ?³ – 2??² + 16 divisible by x + 2?
Answer 16: According to the question,
Let p(x) = x³ – 2mx² + 16, and g(x) = x + 2
⟹ x + 2 = 0
Hence, zero of g(x) = – 2
if p(x) is divisible by g(x), then the remainder of p(−2) should be zero.
Thus, substituting the value of x in p(x), we obtain,
p( – 2) = 0
⟹ ( – 2)³ – 2m( – 2)² + 16 = 0
⟹ 0 – 8 – 8m + 16 = 0
Question 17:If a + b + c = 15 and a² + b² + c² = 83, find the value of a³ + b³ + c³ – 3abc.
Answer 17: We know that,
a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca) ….(i)
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca ….(ii)
Given, a + b + c = 15 and a² + b² + c² = 83
From (ii), we have
152 = 83 + 2(ab + bc + ca)
⇒ 225 – 83 = 2(ab + bc + ca)
⇒ 142/2 = ab + bc + ca
⇒ ab + bc + ca = 71
Now, (i) can be written as
a³ + b³ + c³ – 3abc = (a + b + c)[(a² + b² + c² ) – (ab + bc + ca)]
a³ + b³+ c³ – 3abc = 15 × [83 – 71] = 15 × 12 = 180.
Question 18: Factorise: 27x³+y³+z³–9xyz
Answer 18: The expression27x³+y³+z³–9xyz can be written as (3x)³+y³+z³–3(3x)(y)(z)
27x³+y³+z³–9xyz = (3x)³+y³+z³–3(3x)(y)(z)
We know that x³+y³+³–3xyz = (x+y+z)(x²+y²+z²–xy –yz–zx)
= (3x+y+z)[(3x)²+y²+z²–3xy–yz–3xz]
= (3x+y+z)(9x²+y²+²–3xy–yz–3xz)
Question 19: If (x – 1/x) = 4, then evaluate (x² + 1/x²) and (x⁴ + 1/x⁴).
Answer 19: Given, (x – 1/x) = 4
Squaring both sides, we get,
(x – 1/x)² = 16
⇒ x² – 2.x.1/x + 1/x² = 16
⇒ x² – 2 + 1/x² = 16
⇒ x² + 1/x² = 16 + 2 = 18
∴ (x² + 1/x²) = 18 ….(i)
Again, squaring both sides of (i), we get
(x² + 1/x²)² = 324
⇒ x⁴ + 2.x².1/x² + 1/x⁴= 324
⇒ x⁴ + 2 + 1/x⁴ = 324
⇒ x⁴ + 1/x⁴ = 324 – 2 = 322
∴ (x⁴ + 1/x⁴) = 322.
Question 20: Factorise
Answer 20: The expression 64m³–343n³ can be written as (4m)³–(7n)³
64m³–343n³ =(4m)³–(7n)³
We know that x³–y³ = (x–y)(x²+xy+y²)
64m³–343n³ = (4m)³–(7n)³
= (4m7n)[(4m)²+(4m)(7n)+(7n)²]
= (4m7n)(16m²+28mn+49n² )
Question 21: Find out the values of a and b so that (2x³ + ax² + x + b) has (x + 2) and (2x – 1) as factors.
Answer 21: Let p(x) = 2x³ + ax² + x + b. Then, p( –2) = and p(½) = 0.
p(2) = 2(2)³ + a(2)² + 2 + b = 0
⇒ –16 + 4a – 2 + b = 0 ⇒ 4a + b = 18 ….(i)
p(½) = 2(½)³ + a(½)² + (½) + b = 0
⇒ a + 4b = –3 ….(ii)
On solving (i) and (ii), we get a = 5 and b = –2.
Hence, a = 5 and b = –2.
Question 22: Explain that p – 1 is a factor of p¹⁰ – 1 and p¹¹ – 1.
Answer 22: According to the question,
Let h(p) = ?¹⁰ − 1,and g(p) = ? – 1
zero of g(p) ⇒ g(p) = 0
Therefore, zero of g(x) = 1
We know that,
According to the factor theorem, if g(p) is a factor of h(p), then h(1) should be zero
h(1) = (1) ¹⁰ − 1 = 1 − 1 = 0
⟹ g (p) is a factor of h(p).
Here, we have h(p) = ?¹¹ − 1, g (p) = ? – 1
Putting g (p) = 0 ⟹ ? − 1 = 0 ⟹ ? = 1
As per the factor theorem, if g (p) is a factor of h(p),
Then h(1) = 0
⟹ (1) ¹¹ – 1 = 0
Hence, g(p) = ? – 1 is the factor of h(p) = ? ¹⁰ – 1
Question 23: Examine whether (7 + 3x) is a factor of (3×3 + 7x).
Answer 23: Let p(x) = 3×3 + 7x and g(x) = 7 + 3x. Now g(x) = 0 ⇒ x = –7/3.
By the remainder theorem, p(x) is divided by g(x), and then the remainder is p(–7/3).
Now, p(–7/3) = 3(–7/3)3 + 7(–7/3) = –490/9 ≠ 0.
∴ g(x) is not a factor of p(x).
Question 24:Prove that:
x³+y³+z³–3xyz = (1/2) (x+y+z)[(x–y)²+(y–z)²+(z–x)²]
Answer 24: We know that,
x³+y³+z³−3xyz = (x+y+z)(x²+y²+z²–xy–yz–xz)
⇒ x³+y³+z³–3xyz = (1/2)(x+y+z)[2(x²+y²+z²–xy–yz–xz)]
= (1/2)(x+y+z)(2×2+2y²+²–2xy–2yz–2xz)
= (1/2)(x+y+z)[(x²+y²−2xy)+(y²+z²–2yz)+(x²+z²–2xz)]
= (1/2)(x+y+z)[(x–y)²+(y–z)²+(z–x)²]
Question 25: Find out which of the following polynomials has x – 2 a factor:
(i) 3?² + 6?−24.
(ii) 4?² + ?−2.
Answer 25: (i) According to the question,
Let p(x) =3?² + 6?−24 and g(x) = x – 2
g(x) = x – 2
zero of g(x) ⇒ g(x) = 0
Hence, zero of g(x) = 2
Thus, substituting the value of x in p(x), we get,
p(2) = 3(2)² + 6 (2) – 24
= 12 + 12 – 24
the remainder = zero,
We can derive that,
g(x) = x – 2 is factor of p(x) = 3?² + 6?−24
(ii) According to the question,
Let p(x) = 4?² + ?−2 and g(x) = x – 2
p(2) = 4(2)² + 2−2
Since the remainder ≠ zero,
We can say that,
g(x) = x – 2 is not a factor of p(x) = 4?² + ?−2
Question 26: Factorise x² + 1/x² + 2 – 2x – 2/x.
Answer 26 : x² + 1/x² + 2 – 2x – 2/x = (x² + 1/x² + 2) – 2(x + 1/x)
= (x + 1/x)² – 2(x + 1/x)
= (x + 1/x)(x + 1/x – 2).
Question 27: Factorise
8a³+b³+12a²b+6ab²
Answer 27: The expression, 8a³+b³+12a²b+6ab² can be written as (2a)³+b³+3(2a)²b+3(2a)(b)²
8a³+b³+12a²b+6ab² = (2a)³+b³+3(2a)²b+3(2a)(b)²
= (2a+b)(2a+b)(2a+b)
Here, the identity, (x +y)³ = x³+y³+3xy(x+y) is used.
Question 28: By Remainder Theorem, find out the remainder when p(x) is divided by g(x), where
(i) p(?) = ?³ – 2?² – 4? – 1, g(?) = ? + 1
(ii) p(?) = ?³ – 3?² + 4? + 50, g(?) = ? – 3
(iii) p(?) = 4?³ – 12?² + 14? – 3, g(?) = 2? – 1
(iv) p(?) = ?³ – 6?² + 2? – 4, g(?) = 1 – 3/2 ?
Answer 28: (i) Given p(x) = ?³ – 2?² – 4? – 1 and g(x) = x + 1
Here zero of g(x) = – 1
By applying the remainder theorem
P(x) divided by g(x) = p( – 1)
P ( – 1) = ( – 1)³ – 2 ( – 1)² – 4 ( – 1) – 1 = 0
Therefore, the remainder = 0
(ii) given p(?) = ?³ – 3?² + 4? + 50, g(?) = ? – 3
Here zero of g(x) = 3
By applying the remainder theorem p(x) divided by g(x) = p(3)
p(3) = 3³ – 3 × (3)² + 4 × 3 + 50 = 62
Therefore, the remainder = 62
(iii) p(x) = 4x³ – 12x² + 14x – 3, g(x) = 2x – 1
Here zero of g(x) = ½
By applying the remainder theorem p(x) divided by g(x) = p (½)
P( ½ ) = 4( ½ )³ – 12( ½ )² + 14 ( ½ ) – 3
= 4/8 – 12/4 + 14/2 – 3
= ½ + 1
= 3/2
Hence, the remainder = 3/2
so, zero of g(x) = 2/3
By applying the remainder theorem p(x) divided by g(x) = p(2/3)
p(2/3) = (2/3)³ – 6(2/3)² + 2(2/3) – 4
Therefore, the remainder = – 136/27
Question 29:Factorise x² – 1 – 2a – a².
Answer 29: x² – 1 – 2a – a² = x² – (1 + 2a + a²)
= x² – (1 + a)²
= [x – (1 – a)][x + 1 + a]
= (x – 1 – a)(x + 1 + a)
∴ x² – 1 – 2a – a² = (x – 1 – a)(x + 1 + a).
Question 30:Evaluate the following using suitable identity
Answer 30: We can write 99 as 1000–2
Using identity,(x–y)³ = x³ –y³ –3xy(x–y)
(998)³ =(1000–2)³
=(1000)³ –2³ –(3×1000×2)(1000–2)
= 1000000000–8–6000(1000– 2)
= 1000000000–8 6000000+12000
= 994011992
Question 31: Find the zeroes of the polynomial:
p(?)= (? –2)² −(? + 2)²
Answer 31: p(x) = (? –2)² −(? + 2)²
Zero of the polynomial p(x) = 0
Hence, we get,
⇒ (x–2)² −(x + 2)² = 0
Expanding using the identity, a² – b² = (a – b) (a + b)
⇒ (x – 2 + x + 2) (x – 2 –x – 2) = 0
⇒ 2x ( – 4) = 0
Therefore, the zero of the polynomial = 0
Benefits Of Solving Important Questions Class 9 Mathematics Chapter 2
Consistently solving questions is a vital element of mastering Mathematics. By solving Mathematics Class 9 Chapter 2 important questions, students can get a further understanding of the polynomials chapter.
A few other advantages of solving Important Questions Class 9 Mathematics Chapter 2 are:
 Class 9 Mathematics Chapter 2 important questions provide details about the types of questions that may be expected in the exams, which increases their confidence in achieving a high grade. .
 Studying something once may help you understand the concepts, but through revisions and guided practice make them aware of their mistakes and help to get the best results. Your chances of getting good grades on exams increase as you solve all the questions from our question bank of Important Questions Class 9 Mathematics Chapter 2.
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 By solving our Chapter 2 Class 9 Mathematics important questions, students will understand the question paper pattern.. Practising questions identical to the exam questions would help the students become confident in solving advanced level questions with ease and get 100% results in their exams.
Extramarks leaves no stone unturned to give the best learning material to students while combining fun and learning activities through its own study materials to enhance their learning experience. It provides comprehensive learning solutions for students from Class 1 to Class 12. Our website has abundant resources, along with important questions and solutions. Students can click on the links given below to access some of these resources:
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Q.1 By actual division, find the quotient and the remainder when x 5 + 1 is divided by x 1
Marks: 3 Ans
x 4 + x 3 + x 2 + x + 1 x 1 x 5 + 1 x 5 x 4 + x 4 + 1 x 4 x 3 + x 3 + 1 x 3 x 2 + ¯ x 2 + 1 x 2 x + x + 1 x 1 + 2 ¯ ¯ ¯ ¯ Quotient : x 4 + x 3 + x 2 + x + 1 Remainder : 2
Q.2 Find the value of k if x 5 is a factor of kx 2 + 3x + 7.
Marks: 2 Ans
Zero of x 5 is 5 as x 5 = 0 gives x = 5 . p(x) = kx 2 + 3 x + 7 p ( 5 ) = 0 25 k + 15 + 7 = 0 25 k + 22 = 0 k = 22 25
Q.3 If x + y + z = 6 and xy + yz + zx = 11, then find the value of x 2 + y 2 + z 2 .
Given : x + y + z = 6 and xy + yz + zx = 11 Squaring both sides , we get x + y + z 2 = 6 2 x 2 + y 2 + z 2 + 2 xy + 2 yz + 2 zx = 36 x 2 + y 2 + z 2 + 2 xy + yz + zx = 36 x 2 + y 2 + z 2 + 2 11 = 36 Since xy + yz + zx = 11 x 2 + y 2 + z 2 + 22 = 36 x 2 + y 2 + z 2 = 36 22 = 14
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Faqs (frequently asked questions), 1. what are the four types of polynomials.
The 4 types of polynomials are zero polynomial, linear polynomial, quadratic polynomial, and cubic polynomial.
2. Where can I get important questions for Class 9 Mathematics Chapter 2 online?
On the Extramarks website, you can find all of the important questions for Class 9 Mathematics Chapter 2, along with their answers. On the website, you can also find important questions and NCERT solutions for all classes from 1 to 12.
3. What are the important chapters in Class 9 Mathematics?
The NCERT Mathematics book has 15 chapters. Each chapter is equally important when it comes to learning the fundamentals and taking the test. Additionally, because CBSE does not specify the distribution of marks for each chapter, students are advised to fully study all chapters. Each and every chapter must be completely understood to acquire a good grade in exams.
All the fifteen chapters of CBSE Class 9 Mathematics syllabus are given below:
 Chapter – Number Systems
 Chapter – Polynomials
 Chapter – Coordinate Geometry
 Chapter – Linear Equations In Two Variables
 Chapter – Introduction To Euclid’s Geometry
 Chapter – Lines And Angles
 Chapter – Triangles
 Chapter – Quadrilaterals
 Chapter – Areas Of Parallelograms And Triangles
 Chapter – Circles
 Chapter – Constructions
 Chapter – Heron’s Formula
 Chapter – Surface Areas And Volumes
 Chapter – Statistics
 Chapter – Probability
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 CBSE Maths Important Questions
 Class 9 Maths
Important Questions CBSE Class 9 Maths Chapter 2 Polynomial
Important questions for Class 9 Maths Chapter 2 Polynomials are provided here to help the CBSE students score well in their Class 9 Maths exam. The practice questions given here from polynomials chapter (NCERT) will help the students to create a better understanding of the concepts and, thus, develop their problemsolving skills.
Students can find the CBSE Class 9 Important questions from Chapter 2 Polynomials of the subject Maths here. These questions help students to be familiar with the question types and thus face the exam more confidently.
Also Check:
 Important 2 Marks Questions for CBSE 9th Maths
 Important 3 Marks Questions for CBSE 9th Maths
 Important 4 Marks Questions for CBSE 9th Maths
Important Polynomials Questions For Class 9 Chapter 2 (With Solutions)
Some important questions from polynomials are given below with solutions. These questions will help the 9th class students to get acquainted with a wide variety of questions and develop the confidence to solve polynomial questions more efficiently.
1. Give an example of a monomial and a binomial having degrees of 82 and 99, respectively.
An example of a monomial having a degree of 82 = x 82
An example of a binomial having a degree of 99 = x 99 + x
2. Compute the value of 9x 2 + 4y 2 if xy = 6 and 3x + 2y = 12.
Consider the equation 3x + 2y = 12
(3x + 2y) 2 = 12 2
=> 9x 2 + 12xy + 4y 2 = 144
=>9x 2 + 4y 2 = 144 – 12xy
9x 2 + 4y 2 = 144 – 72
Thus, the value of 9x 2 + 4y 2 = 72
3. Find the value of the polynomial 5x – 4x 2 + 3 at x = 2 and x = –1.
Let the polynomial be f(x) = 5x – 4x 2 + 3
f(2) = 5(2) – 4(2) 2 + 3
Or, the value of the polynomial 5x – 4x 2 + 3 at x = 2 is 3.
f(–1) = 5(–1) – 4(–1) 2 + 3
=> f(–1) = –5 –4 + 3 = 6
The value of the polynomial 5x – 4x 2 + 3 at x = 1 is 6.
4. Calculate the perimeter of a rectangle whose area is 25x 2 – 35x + 12.
Area of rectangle = 25x 2 – 35x + 12
We know, area of rectangle = length × breadth
So, by factoring 25x 2 – 35x + 12, the length and breadth can be obtained.
25x 2 – 35x + 12 = 25x 2 – 15x – 20x + 12
=> 25x 2 – 35x + 12 = 5x(5x – 3) – 4(5x – 3)
=> 25x 2 – 35x + 12 = (5x – 3)(5x – 4)
So, the length and breadth are (5x – 3)(5x – 4).
Now, perimeter = 2(length + breadth)
= 2(5x – 3 + 5x – 4) = 2(10x – 7) = 20x – 14
So, the perimeter = 20x – 14
5. Find the value of x 3 + y 3 + z 3 – 3xyz if x 2 + y 2 + z 2 = 83 and x + y + z = 15
Consider the equation x + y + z = 15
From algebraic identities, we know that (a + b + c) 2 = a 2 + b 2 + c 2 + 2(ab + bc + ca)
(x + y + z) 2 = x 2 + y 2 + z 2 + 2(xy + yz + xz)
From the question, x 2 + y 2 + z 2 = 83 and x + y + z = 15
15 2 = 83 + 2(xy + yz + xz)
=> 225 – 83 = 2(xy + yz + xz)
Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca),
x 3 + y 3 + z 3 – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz))
So, x 3 + y 3 + z 3 – 3xyz = 15(83 – 71)
=> x 3 + y 3 + z 3 – 3xyz = 15 × 12
Or, x 3 + y 3 + z 3 – 3xyz = 180
6. If a + b + c = 15 and a 2 + b 2 + c 2 = 83, find the value of a 3 + b 3 + c 3 – 3abc.
a 3 + b 3 + c 3 – 3abc = (a + b + c)(a 2 + b 2 + c 2 – ab – bc – ca) ….(i)
(a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca ….(ii)
Given, a + b + c = 15 and a 2 + b 2 + c 2 = 83
15 2 = 83 + 2(ab + bc + ca)
a 3 + b 3 + c 3 – 3abc = (a + b + c)[(a 2 + b 2 + c 2 ) – (ab + bc + ca)]
a 3 + b 3 + c 3 – 3abc = 15 × [83 – 71] = 15 × 12 = 180.
7. If (x – 1/x) = 4, then evaluate (x 2 + 1/x 2 ) and (x 4 + 1/x 4 ).
Given, (x – 1/x) = 4
Squaring both sides we get,
(x – 1/x) 2 = 16
⇒ x 2 – 2.x.1/x + 1/x 2 = 16
⇒ x 2 – 2 + 1/x 2 = 16
⇒ x 2 + 1/x 2 = 16 + 2 = 18
∴ (x 2 + 1/x 2 ) = 18 ….(i)
(x 2 + 1/x 2 ) 2 = 324
⇒ x 4 + 2.x 2 .1/x 2 + 1/x 4 = 324
⇒ x 4 + 2 + 1/x 4 = 324
⇒ x 4 + 1/x 4 = 324 – 2 = 322
∴ (x 4 + 1/x 4 ) = 322.
8. Find the values of a and b so that (2x 3 + ax 2 + x + b) has (x + 2) and (2x – 1) as factors.
Let p(x) = 2x 3 + ax 2 + x + b. Then, p( –2) = and p(½) = 0.
p(2) = 2(2) 3 + a(2) 2 + 2 + b = 0
p(½) = 2(½) 3 + a(½) 2 + (½) + b = 0
9. Check whether (7 + 3x) is a factor of (3x 3 + 7x).
Let p(x) = 3x 3 + 7x and g(x) = 7 + 3x. Now g(x) = 0 ⇒ x = –7/3.
By the remainder theorem , we know that when p(x) is divided by g(x) then the remainder is p(–7/3).
Now, p(–7/3) = 3(–7/3) 3 + 7(–7/3) = –490/9 ≠ 0.
10. Factorise x 2 + 1/x 2 + 2 – 2x – 2/x.
Solution:
x 2 + 1/x 2 + 2 – 2x – 2/x = (x 2 + 1/x 2 + 2) – 2(x + 1/x)
= (x + 1/x) 2 – 2(x + 1/x)
11. Factorise x 2 – 1 – 2a – a 2 .
x 2 – 1 – 2a – a 2 = x 2 – (1 + 2a + a 2 )
= x 2 – (1 + a) 2
∴ x 2 – 1 – 2a – a 2 = (x – 1 – a)(x + 1 + a).
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its very nice app
tough please make the solution a bit easy
I thank byjus for providing these important questions
Nice questions but little bit difficult make it easy to solve
I don’t got question 5, step x^3+y^3+z^33xyz=15(8371) Why we subtracted 8371??
We evaluated the value of xy + yz + xz, initially, which is equal to 71. Since, x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz)) [By algebraic identities] And x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71 So, if we substitute the values, we get: x^3 + y^3 + z^3 – 3xyz = 15(83 – 71) Please go through the complete solution for the answer.
The expression is x^3+y^3+z^3 = (x+y+z)(x^2+y^2+z^2xyyzxz)
= (x+y+z){(x^2+y^2+z^2)(xy+yz+sa)}
x^2 +y^2 + z^2 = 83, xy+yz+sa = 71 This is why 83 71 is done
x + y + z = 15, x² + y² + z² = 83 And xy + yz + xz = 71 Now, Acc. To the question, x3 + y3 + z3 – 3xyz which is equal to = (x + y + z)(x² + y² + z² – (xy + yz + xz)) And these are given above So we subtract 83 – 71
because the formula is x^3+y^3+z^33xyz = (x+y+z)(x^2 + y^2 + z^2xyyzzx) and we know x+y+z= 15 and x^2 + y^2 + z^2 = 83 and xy+yz+zx = 71 and xyyzzx = 71 so we have putted the values on formula 15(8371)
we found the value of xy+yz+zx so in the above step we take ‘‘ as common and then xy+yz+zx so it will 71 then it is 15*12=180
This is so nice we can learn easily
wow! nice app
Thank you so much for these questions
Nice questions and its also easy to check since the solutions are given
Very helpful during online education
maths in byjus is easy. thank you for everything
Please add some more questions. There should be at least 10 questions
very good question for the exam. thank you
Nice app but difficult to do the question plz make it
Very easy to solve
Very nice app and there question is very useful to me, there question is very interesting .
These are the most basic and the most common questions for exams. Thank you Byjus app for making studies easier than ever.
These are the most basic , the most common and the most tough questions for exams. Thank you Byjus app for making studies easier than ever.
thank you for providing these important questions
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CBSE Case Study Questions Class 9 Maths Chapter 2 Polynomials PDF Download
CBSE Case Study Questions Class 9 Maths Chapter 2 Polynomials PDF Download are very important to solve for your exam. Class 9 Maths Chapter 2 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Case Study Questions Class 9 Maths Chapter 2 Polynomials
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Polynomials Case Study Questions With Answers
Case study questions class 9 maths chapter 2.
Case Study/PassageBased Questions
Case Study 1. Ankur and Ranjan start a new business together. The amount invested by both partners together is given by the polynomial p(x) = 4x 2 + 12x + 5, which is the product of their individual shares.
Coefficient of x 2 in the given polynomial is (a) 2 (b) 3 (c) 4 (d) 12
Answer: (c) 4
Total amount invested by both, if x = 1000 is (a) 301506 (b)370561 (c) 4012005 (d)490621
Answer: (c) 4012005
The shares of Ankur and Ranjan invested individually are (a) (2x + 1),(2x + 5)(b) (2x + 3),(x + 1) (c) (x + 1),(x + 3) (d) None of these
Answer: (a) (2x + 1),(2x + 5)
Name the polynomial of amounts invested by each partner. (a) Cubic (b) Quadratic (c) Linear (d) None of these
Answer: (c) Linear
Find the value of x, if the total amount invested is equal to 0. (a) –1/2 (b) –5/2 (c) Both (a) and (b) (d) None of these
Answer: (c) Both (a) and (b)
Case Study 2. One day, the principal of a particular school visited the classroom. The class teacher was teaching the concept of a polynomial to students. He was very much impressed by her way of teaching. To check, whether the students also understand the concept taught by her or not, he asked various questions to students. Some of them are given below. Answer them
Which one of the following is not a polynomial? (a) 4x 2 + 2x – 1 (b) y+3/y (c) x 3 – 1 (d) y 2 + 5y + 1
Answer: (b) y+3/y
The polynomial of the type ax 2 + bx + c, a = 0 is called (a) Linear polynomial (b) Quadratic polynomial (c) Cubic polynomial (d) Biquadratic polynomial
Answer: (a) Linear polynomial
The value of k, if (x – 1) is a factor of 4x 3 + 3x 2 – 4x + k, is (a) 1 (b) –2 (c) –3 (d) 3
Answer: (c) –3
If x + 2 is the factor of x 3 – 2ax 2 + 16, then value of a is (a) –7 (b) 1 (c) –1 (d) 7
Answer: (b) 1
The number of zeroes of the polynomial x 2 + 4x + 2 is (a) 1 (b) 2 (c) 3 (d) 4
Answer: (b) 2
Case Study 3. Amit and Rahul are friends who love collecting stamps. They decide to start a stamp collection club and contribute funds to purchase new stamps. They both invest a certain amount of money in the club. Let’s represent Amit’s investment by the polynomial A(x) = 3x^2 + 2x + 1 and Rahul’s investment by the polynomial R(x) = 2x^2 – 5x + 3. The sum of their investments is represented by the polynomial S(x), which is the sum of A(x) and R(x).
Q1. What is the coefficient of x^2 in Amit’s investment polynomial A(x)? (a) 3 (b) 2 (c) 1 (d) 0
Answer: (a) 3
Q2. What is the constant term in Rahul’s investment polynomial R(x)? (a) 2 (b) 5 (c) 3 (d) 0
Answer: (c) 6
Q3. What is the degree of the polynomial S(x), representing the sum of their investments? (a) 4 (b) 3 (c) 2 (d) 1
Answer: (c) 2
Q4. What is the coefficient of x in the polynomial S(x)? (a) 7 (b) 3 (c) 0 (d) 5
Answer: (b) 3
Q5. What is the sum of their investments, represented by the polynomial S(x)? (a) 5x^2 + 7x + 4 (b) 5x^2 – 3x + 4 (c) 5x^2 – 3x + 5 (d) 5x^2 + 7x + 5
Answer: (b) 5x^2 – 3x + 4
Case Study 4. A school is organizing a fundraising event to support a local charity. The students are divided into three groups: Group A, Group B, and Group C. Each group is responsible for collecting donations from different areas of the town.
Group A consists of 30 students and each student is expected to collect ‘x’ amount of money. The polynomial representing the total amount collected by Group A is given as A(x) = 2x^2 + 5x + 10.
Group B consists of 20 students and each student is expected to collect ‘y’ amount of money. The polynomial representing the total amount collected by Group B is given as B(y) = 3y^2 – 4y + 7.
Group C consists of 40 students and each student is expected to collect ‘z’ amount of money. The polynomial representing the total amount collected by Group C is given as C(z) = 4z^2 + 3z – 2.
Q1. What is the coefficient of x in the polynomial A(x)? (a) 2 (b) 5 (c) 10 (d) 0
Answer: (b) 5
Q2. What is the degree of the polynomial B(y)? (a) 2 (b) 3 (c) 4 (d) 1
Answer: (b) 3
Q3. What is the constant term in the polynomial C(z)? (a) 4 (b) 3 (c) 2 (d) 0
Answer: (c) 2
Q4. What is the sum of the coefficients of the polynomial A(x)? (a) 2 (b) 5 (c) 10 (d) 17
Answer: (c) 10
Q5. What is the total number of students in all three groups combined? (a) 30 (b) 20 (c) 40 (d) 90
Answer: (c) 40
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Chapter 2 Class 9 Polynomials
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Solutions to all NCERT Exercise Questions and Examples of Chapter 2 Class 9 Polynomials are provided free at Teachoo. Answers to each and every question is explained in an easy to understand way, with videos of all the questions.
In this chapter, we will learn
 What is a Polynomial
 What are Polynomials in One Variable
 What are monomials, binomials, trinomials
 What is the Degree of a Polynomial
 What are Linear, Quadratic and Cubic Polynomials
 What is zero of a polynomial (or root of a polynomial)
 Finding zeroes of a polynomial
 Dividing Polynomials and finding remainder
 Finding Remainder using Remainder Theorem
 Checking if it is a factor or not
 Factorising Quadratic and Cubic Polynomials using Factor Theorem
 Solving questions using Algebra Identities (Check full list of Algebra Formulas )
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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12
NCERT Exemplar Class 9 Maths Chapter 2 Polynomials
April 10, 2019 by Bhagya
NCERT Exemplar Class 9 Maths Chapter 2 Polynomials are part of NCERT Exemplar Class 9 Maths . Here we have given NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials.
NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials
Exercise 2.1
Question 2. √2 is a polynomial of degree (a) 2 (b) 0 (c) 1 (d)½ Solution: (b) √2 = √2x°. Hence, √2 is a polynomial of degree 0, because exponent of x is 0.
Question 3. Degree of the polynomial 4x 4 + 0x 3 + 0x 5 + 5x + 7 is (a) 4 (b) 5 (c) 3 (d) 7 Solution: (a) 4 4x 4 + 0x 3 + 0x 5 + 5x + 7 = 4x 4 + 5x + 7 As we know that the degree of a polynomial is equal to the highest power of variable x. Here, the highest power of x is 4. Therefore, the degree of the given polynomial is 4.
Question 4. Degree of the zero polynomial is (A) 0 (B) 1 (C) Any natural number (D) Not defined Solution: (D): In zero polynomial, the coefficient of any power of variable is zero i.e., 0x², 0x 5 etc. Therefore, we can not exactly determine the highest power of variable, hence cannot define the degree of zero polynomial.
Question 5. If p (x) = x 2 – 2√2x + 1, then p (2√2) is equal to (a) 0 (b) 1 (c) 4√2 (d) 8 √2 +1 Solution: (b) Given, p(x) = x 2 – 2√2x + 1 …(i) On putting x = 2√2 in Eq. (i), we get P(2√2) = (2√2) 2 – (2√2)(2√2) + 1 = 8 – 8 + 1 = 1
Question 6. The value of the polynomial 5x – 4x 2 + 3, when x = – 1 is (a)6 (b) 6 (c) 2 (d) 2 Solution: (a) Let p (x) = 5x – 4x 2 + 3 …(i) On putting x = 1 in Eq. (i), we get p(1) = 5(1) 4(1) 2 + 3= 5 – 4 + 3 = 6
Question 7. If p (x) = x + 3, then p(x) + p( x) is equal to (a) 3 (b) 2x (c) 0 (d) 6 Solution: (d) Given p(x) = x + 3, put x = x in the given equation, we get p(x) = x + 3 Now, p(x) + p(x) = x + 3 + (x) + 3 = 6
Question 8. Zero of the zero polynomial is (a) 0 (b) 1 (c) any real number (d) not defined Solution: (c) Zero of the zero polynomial is any real number. e.g., Let us consider zero polynomial be 0(x – k), where k is a real number For determining the zero, put x – k = 0 ⇒ x = k Hence, zero of the zero polynomial be any real number.
Question 9. Zero of the polynomial p(x) = 2x + 5 is (a) 2/5 (b) 5/2 (c) 2/5 (d) 5/2 Solution: (b) Given, p(x) = 2x + 5 For zero of the polynomial, put p(x) = 0 ∴ 2x + 5 = 0 ⇒ 5/2 Hence, zero of the polynomial p(x) is 5/2.
Question 10. One of the zeroes of the polynomial 2x 2 + 7x – 4 is (a) 2 (b) ½ (c) 1 (d) 2 Solution: (b) Let p (x) = 2x 2 + 7x – 4 = 2x 2 + 8x – x – 4 [by splitting middle term] = 2x(x+ 4)1(x + 4) = (2x1)(x+ 4) For zeroes of p(x), put p(x) = 0 ⇒ (2x 1) (x + 4) = 0 ⇒ 2x – 1 = 0 and x + 4 = 0 ⇒ x = ½ and x = 4 Hence, one of the zeroes of the polynomial p(x) is ½.
Question 11. If x 51 + 51 is divided by x + 1, then the remainder is (a) 0 (b) 1 (c) 49 (d) 50 Solution: (d) Let p(x) = x 51 + 51 . …(i) When we divide p(x) by x+1, we get the remainder p(1) On putting x = 1 in Eq. (i), we get p(1) = (1) 51 + 51 = 1 + 51 = 50 Hence, the remainder is 50.
Question 12. If x + 1 is a factor of the polynomial 2x 2 + kx, then the value of k is (a) 3 (b) 4 (c) 2 (d)2 Solution: (c) Let p(x) = 2x 2 + kx Since, (x + 1) is a factor of p(x), then p(1)=0 2(1)2 + k(1) = 0 ⇒ 2 – k = 0 ⇒ k = 2 Hence, the value of k is 2.
Question 13. x + 1 is a factor of the polynomial (a) x 3 + x 2 – x + 1 (b) x 3 + x 2 + x + 1 (c) x 4 + x 3 + x 2 + 1 (d) x 4 + 3x 3 + 3x 2 + x + 1 Solution: (b) Let assume (x + 1) is a factor of x 3 + x 2 + x + 1. So, x = 1 is zero of x 3 + x 2 + x + 1 (1) 3 + (1) 2 + (1) + 1 = 0 ⇒ 1 + 1 – 1 + 1 = 0 ⇒ 0 = 0 Hence, our assumption is true.
Question 14. One of the factors of (25x 2 – 1) + (1 + 5x) 2 is (a) 5 + x (b) 5 – x (c) 5x 1 (d) 10x Solution: (d) Now, (25x 2 1) + (1 + 5x) 2 = 25x 2 1 + 1 + 25x 2 + 10x [using identity, (a + b) 2 = a 2 + b 2 + 2ab] = 50x 2 + 10x = 10x (5x + 1) Hence, one of the factor of given polynomial is 10x.
Question 15. The value of 249 2 – 248 2 is (a) 1 2 (b) 477 (c) 487 (d) 497 Solution: (d) Now, 249 2 – 248 2 = (249 + 248) (249 – 248) [using identity, a 2 – b 2 = (a – b)(a + b)] = 497 × 1 = 497.
Question 16. The factorization of 4x 2 + 8x+ 3 is (a) (x + 1) (x + 3) (b) (2x + 1) (2x + 3) (c) (2x + 2) (2x + 5) (d) (2x – 1) (2x – 3) Solution: (b) Now, 4x 2 + 8x + 3= 4x 2 + 6x + 2x + 3 [by splitting middle term] = 2x(2x + 3) + 1 (2x + 3) = (2x + 3) (2x + 1)
Question 17. Which of the following is a factor of (x+ y) 3 – (x 3 + y 3 )? (a) x 2 + y 2 + 2 xy (b) x 2 + y 2 – xy (c) xy 2 (d) 3xy Solution: (d) Now, (x+ y)3 – (x 3 + y 3 ) = (x + y) – (x + y)(x 2 – xy + y 2 ) [using identity, a 3 + b 3 = (a + b)(a 2 – ab + b 2 )] = (x + y)[(x + y) 2 (x 2 – xy + y 2 )] = (x+ y)(x 2 + y 2 + 2xy – x 2 + xy – y 2 ) [using identity, (a + b) 2 = a 2 + b 2 + 2 ab)] = (x + y) (3xy) Hence, one of the factor of given polynomial is 3xy.
Question 18. The coefficient of x in the expansion of (x + 3) 3 is (a) 1 (b) 9 (c) 18 (d) 27 Solution: (d) Now, (x + 3) 3 = x 3 + 3 3 + 3x (3)(x + 3) [using identity, (a + b) 3 = a 3 + b 3 + 3ab (a + b)] = x 3 + 27 + 9x (x + 3) = x 3 + 27 + 9x 2 + 27x Hence, the coefficient of x in (x + 3) 3 is 27.
Question 21. If a + b + c =0, then a 3 + b 3 + c 3 is equal to (a) 0 (b) abc (c) 3abc (d) 2abc Solution: (d) Now, a 3 + b 3 + c 3 = (a + b + c) (a 2 + b 2 + c 2 – ab – be – ca) + 3abc [using identity, a 3 + b 3 + c 3 – 3 abc = (a + b + c)(a 2 + b 2 + c 2 – ab – be – ca)] = 0 + 3abc [∴ a + b + c = 0, given] a 3 + b 3 + c 3 = 3abc
Exercise 2.2
(ii) Polynomial Each exponent of the variable x is a whole number.
(iii) Not polynomial Because exponent of the variable x is 1/2, which is not a whole number.
(iv) Polynomial Because each exponent of the variable x is a whole number.
(v) Not polynomial Because one of the exponents of the variable x is 1, which is not a whole number.
(vi) Not polynomial Because given expression is a rational expression, thus, not a polynomial.
(vii) Polynomial Because each exponent of the variable a is a whole number.
Question 2. Write whether the following statements are True or False. Justify your answer. (i) A binomial can have atmost two terms (ii) Every polynomial is a binomial (iii) A binomial may have degree 5 (iv) Zero of a polynomial is always 0 (v) A polynomial cannot have more than one zero (vi) The degree of the sum of two polynomials each of degree 5 is always 5. Solution: (i) False Because a binomial has exactly two terms.
(ii) False Because every polynomial is not a binomial. e.g., (a)x² + 4x + 3 [polynomial but not a binomial] (b) x² + 5 [polynomial and also a binomial]
(iii) True Because a binomial is a polynomial whose degree is a whole number which is greater than or equal to one. Therefore, a binomial may have degree 5.
(iv) False Because zero of a polynomial can be any real number e.g., for p(x) = x – 1, zero of p(x) is 1, which is a real number.
(v) False Because a polynomial can have any number of zeroes. It depends upon the degree of the polynomial. e.g. for p(x) = x² – 4, degree is 2, so it has two zeroes i.e., 2 and 2.
(vi) False Because the sum of any two polynomials of same degree has not always same degree. e.g., Let f(x) = x 5 + 2 and g(x) = x 5 + 2x 2 ∴ Sum of two polynomials, f(x) + g(x) = x 5 + 2 + (x 5 + 2x 2 ) = 2x 2 + 2, which is not a polynomial of degree 5.
Exercise 2.3
Question 1. Classify the following polynomials as polynomials in one variable, two variables etc. (i) x 2 + x + 1 (ii) y 3 – 5y (iii) xy + yz + zx (iv) x 2 – Zxy + y 2 + 1 Solution: (i) Polynomial x 2 + x + 1 is a one variable polynomial, because it contains only one variable i.e., x. (ii) Polynomial y 3 – 5y is a one variable polynomial, because it contains only one variable i.e., y. (iii) Polynomial xy + yz + zx is a three variables polynomial, because it contains three variables x, y and z. (iv) Polynomial x 2 – 2xy + y 2 + 1 is a two variables polynomial, because it contains two variables x and y.
Question 2. Determine the degree of each of the following polynomials. (i) 2x – 1 (ii) 10 (iii) x 3 – 9x + 3x 5 (iv) y 3 (1 – y 4 ) Solution: (i) Degree of polynomial 2x – 1 is 1, Because the maximum exponent of x is 1. (ii) Degree of polynomial – 10 or – 10x° is 0, because the exponent of x is 0. (iii) Degree of polynomial x3 – 9x + 3x 5 is 5, because the maximum exponent of x is 5. (iv) Degree of polynomial y 3 (1 – y 4 ) or y 3 – y 7 is 7, because the maximum exponent of y is 7.
(ii) The coefficient of x 3 in given polynomial is 1/5
(iii) The coefficient of x 6 in given polynomial is 1.
(iv) The constant term in given polynomial is 1/5
(iii) We have, (x – 1) (3x – 4) = 3x 2 – 7x + 4 ∴ Coefficient of x² in 3x² – 7x + 4 is 3.
(iv) We have, (2x – 5) (2x² – 3x + 1) = 2x (2x² – 3x + 1) – 5(2x² – 3x + 1) = 4x³ – 6x² + 2x – 10x² + 15x – 5 = 4x³ – 16x² + 17x – 5 Coefficient of x² in 4x³ – 16x² + 17x – 5 is 16.
Question 5. Classify the following as constant, linear, quadratic and cubic polynomials: (i) 2 – x² + x³ (ii) 3x³ (iii) 5t – √7 (iv) 4 – 5y² (v) 3 (vi) 2 + x (vii) y³ – y (viii) 1 + x + x² (ix) t² (x) √2x – 1 Solution: (i) Polynomial 2 – x² + x³ is a cubic polynomial, because its degree is 3. (ii) Polynomial 3x³ is a cubic polynomial, because its degree is 3. (iii) Polynomial 5t – √7 is a linear polynomial, because its degree is 1. (iv) Polynomial 4 – 5y² is a quadratic polynomial, because its degree is 2. (v) Polynomial 3 = 3x° is a constant polynomial, because its degree is 0. (vi) Polynomial 2 + x is a linear polynomial, because its degree is 1. (vii) Polynomial y³ – y is a cubic polynomial, because its degree is 3. (viii) Polynomial 1 + x + x² is a quadratic polynomial, because its degree is 2. (ix) Polynomial t² is a quadratic polynomial, because its degree is 2. (x) Polynomial √2x – 1 is a linear polynomial, because its degree is 1.
Question 6. Give an example of a polynomial, which is (i) monomial of degree 1. (ii) binomial of degree 20. (iii) trinomial of degree 2. Solution: (i) The example of monomial of degree 1 is 3x. (ii) The example of binomial of degree 20 is 3x 20 + x 10 (iii) The example of trinomial of degree 2 is x 2 – 4x + 3
Question 7. Find the value of the polynomial 3x 3 – 4x 2 + 7x – 5, when x = 3 and also when x = 3. Solution: Let p(x) =3x 3 – 4x 2 + 7x – 5 At x= 3, p(3) = 3(3) 3 – 4(3) 2 + 7(3) – 5 = 81 – 36 + 21 – 5 ∴ P( 3) = 61 At x = 3, p(3)= 3(3) 3 – 4(3) 2 + 7(3) – 5 = 81 – 36 – 21 – 5 = 143 ∴ p(3) = 143 Hence, the value of the given polynomial at x = 3 and x = 3 are 61 and 143, respectively.
Question 9. Find p(0), p( 1) and p(2) for the following polynomials (i) p(x) = 10x – 4x 2 – 3 (ii) p(y) = (y + 2)(y – 2) Solution: (i) Given, polynomial is p(x) = 10x – 4x 2 – 3 On putting x = 0, 1 and – 2, respectively in Eq. (i), we get p(0) = 10(0) – 4(0) 2 – 3 = 0 – 0 – 3 = 3 p(1) = 10 (1) – 4 (1 ) 2 – 3 = 10 – 4 – 3= 10 – 7 = 3 and p(2) =10 (2) 4 (2) 2 – 3 = 20 4 × 4 – 3 = 20 – 16 – 3 = 39 Hence, the values of p(0), p(1) and p(2) are respectively, 3, 3 and – 39.
(ii) Given, polynomial isp(y) = (y+2)(y2) On putting y =0,1 and 2, respectively in Eq. (i), we get p(0) = (0+2)(02) = 4 p(1) = (1 + 2)(12) = 3 x (1) = 3 and p(2) = (2 + 2)(2 2) = 0 (4) = 0 Hence, the values of p(0), p(1) and p(2) are respectively, 4, 3 and 0.
Question 10. Verify whether the following are true or false. (i) 3 is a zero of at – 3 (ii) 1/3 is a zero of 3x + 1 (iii) 4/5 is a zero of 4 – 5y (iv) 0 and 2 are the zeroes of t 2 – 2t (v) 3 is a zero of y 2 + y – 6 Solution: (i) False Put x – 3 = 0 ⇒ x = 3 Hence, zero of x – 3 is 3.
(ii) True Put 3x + 1 = 0 ⇒ x = 1/3 Hence, zero of 3x + 1 is 1/5.
(iii) False Put 4 – 5y = 0 ⇒ y = 4/5 Hence, zero of 4 – 5y is 4/5.
(iv) True Put t² – 2t = 0 ⇒ t(t – 2) = 0 ⇒ t = 0 and t – 2 = 0 ⇒ t = 0 and t = 2 Hence, the zeroes of t² – 2t are 0 and 2.
(v) True Put y² + y – 6 = 0 ⇒ y² + 3y – 2y – 6 = 0 ⇒ y(y + 3) – 2(y + 3) = 0 = (y2)(y + 3) = 0 ⇒ y – 2 = 0 and y + 3 = 0 ⇒ y = 2 and y = 3 Hence, the zeroes of y² + y – 6 are 2 and – 3.
Question 11. Find the zeroes of the polynomial in each of the following, (i) p(x)= x – 4 (ii) g(x)= 3 – 6x (iii) q(x) = 2x – 7 (iv) h(y) = 2y Solution: (i) Given, polynomial is p(x) = x – 4 For zero of polynomial, put p(x) = 0 ∴ x – 4 = 0 ⇒ x = 4 Hence, zero of polynomial is 4.
(ii) Given, polynomial is g(x) = 3 – 6x For zero of polynomial, put g(x) = 0 ∴ 3 – 6x = 0 ⇒ x = 3/6 = 1/2. Hence, zero of polynomial is X
(iii) Given, polynomial is q(x) = 2x – 7 For zero of polynomial, put q(x) = 0 ∴ 2x – 7 = 0 ⇒ 2x = 7 ⇒ x = 7/2 Hence, zero of polynomial q(x) is 7/2
(iv) Given polynomial h(y) = 2 y For zero of polynomial, put h(y) = 0 ∴ 2y = 0 ⇒ y = 0 Hence, the zero of polynomial h(y) is 0.
Question 12. Find the zeroes of the polynomial p(x) = (x – 2) 2 – (x + 2) 2 . Solution: Given, polynomial is p(x) = (x – 2) 2 – (x + 2) 2 For zeroes of polynomial, put p(x) = 0 ∴ (x – 2) 2 – (x + 2) 2 = 0 ⇒ (x – 2 + x + 2)(x – 2 – x – 2) = 0 [using identity, a 2 – b 2 = (a – b)(a + b)] ⇒ (2x)(4) = 0 ⇒ 8x = 0 ⇒ x = 0 Therefore zero of the polynomial is p(x) is 0.
Question 14. By remainder theorem, find the remainder when p(x) is divided by g(x) (i) p(x) = x 3 – 2x 2 – 4x – 1, g(x) = x + 1 (ii) p(x) = x 3 – 3x 2 + 4x + 50, g(x) = x – 3 (iii) p(x) = x 3 – 12x 2 + 14x – 3, g(x) = 2x – 1 – 1 (iv) p(x) = x 3 – 6x 2 + 2x – 4, g(x) = 1 – (3/2) x Solution: (i) We have, p(x) = x³ – 2x² – 4x – 1 and g(x) = x + 1 Here, zero of g(x) is 1. When we divide p(x) by g(x) using remainder theorem, we get the remainder p(1) ∴ p(1) = (1)³ – 2(1)² – 4(1) 1 = 1 – 2 + 4 – 1 = 0 Therefore, remainder is 0.
(ii) We have, p(x) = x³ – 3x² + 4x + 50 and g(x) = x – 3 Here, zero of g(x) is 3. When we divide p(x) by g(x) using remainder theorem, we get the remainder p(3) ∴ p(3) = (3)³ – 3(3)² + 4(3) + 50 = 27 – 27 + 12 + 50 = 62 Therefore, remainder is 62.
Question 15. Check whether p(x) is a multiple of g(x) or not (i) p(x) = x 3 – 5x 2 + 4x – 3, g(x) = x – 2. (ii) p(x) = 2x 3 – 11x 2 – 4x + 5, g(x) = 2x + 1 Solution: (i) We have, g(x) = x – 2 Here, zero of g(x) is 2. Now, p(x) = x 3 – 5x 2 + 4x – 3 ∴ p(2) = (2) 3 – 5(2) 2 + 4(2) – 3 = 8 – 20 + 8 – 3 = – 7 Since, remainder ≠ 0, then p(x) is not a multiple of g(x).
Since, remainder ≠ 0, then p(x) is not a multiple of g(x).
Question 16. Show that, (i) x + 3 is a factor of 69 + 11x – x 2 + x 3 (ii) 2x – 3 is a factor of x + 2x 3 9x 2 +12 Solution: (i) Let p(x) = x 3 – x 2 + 11x + 69 x + 3 is a factor of p(x) if p(3) = 0 Now, p(3) = (3) 3 – (3) 2 + 11(3) + 69 = – 27 – 9 – 33 + 69 = 0 Therefore, (x + 3) is a factor of p(x).
Question 17. Determine which of the following polynomial has x – 2 a factor (i) 3x 2 + 6x – 24 (ii) 4x 2 + x – 2 Solution: (i) Let p(x) = 3x 2 + 6x – 24 … (1) Substituting x = 2 in (1), we get p(2) = 3(2) 2 + 6(2) – 24 = 12 + 12 – 24 = 0 Hence, x – 2 is a factor of p(x).
(ii)Let p(x) = 4x 2 + x – 2 … (2) Substituting x = 2 in (2), we get p(2) = 4(2) 2 + 2 – 2 = 16 ≠ 0 Hence, x – 2 is not a factor of p(x).
Question 18. Show that p1 is a factor of p 10 1 and also of p 11 1. Solution: Let g (p) = p 10 1 …(1) and h(p) = p 11 1 …(2) On putting p = 1 in Eq. (i), we get g(1) = 1 10 1 = 1 – 1 = 0 Hence, p1 is a factor of g(p). Again, putting p = 1 in Eq. (2), we get h (1) = (1) 11 1 = 1 – 1 = 0 Hence, p 1 is a factor of h(p).
Question 19. For what value of m is x 3 2mx 2 +16 divisible by x + 2? Solution: Let p(x) = x 3 – 2mx 2 + 16 Since, p(x) is divisible by (x+2), then remainder = 0 P(2) = 0 ⇒ (2) 3 – 2m(2) 2 + 16 = 0 ⇒ 8 – 8m + 16 = 0 ⇒ 8m = 8 m = 1 Hence, the value of m is 1.
Question 20. If x + 2a is a factor of a 5 – 4a 2 x 3 + 2x + 2a + 3, then find the value of a. Solution: Let p(x) =a 5 – 4a 2 x 3 + 2x + 2a + 3 Since, x + 2a is a factor of p(x), then put p(2a) = 0 ∴ (2a) 5 – 4a 2 (2a) 3 + 2(2a) + 2a + 3 = 0 ⇒ 32a 5 + 32a 5 – 4a + 2a + 3 = 0 ⇒ 2a + 3 = 0 2a = 3 a = 3/2. Hence, the value of a is 3/2.
Question 22. If x + 1 is a factor of ax 3 + x 2 – 2x + 4o – 9, find the value of a. Solution: Let p(x) = ax 3 + x 2 – 2x + 4a – 9 Since, x + 1 is a factor of p(x), then p(1) = 0 a( 1) 3 + ( 1) 2 – 2(1) + 4a – 9 = 0 ⇒ a + 1 + 2 + 4a – 9 = 0 ⇒ 3a = 6 ⇒ a = 2
Question 23. Factorise: (i) x 2 + 9x + 18 (ii) 6x 2 + 7x – 3 (iii) 2x 2 – 7x – 15 (iv) 84 – 2r – 2r 2 Solution: (i) We have, x 2 + 9x +18 = x 2 + 6x + 3x +18 = x(x + 6) + 3(x + 6) = (x + 3) (x + 6) We have, 6x 2 + 7x – 3 = 6x 2 + 9x – 2x – 3 = 3x(2x + 3) – 1(2x + 3) = (3x – 1)(2x + 3) We have, 2x 2 – 7x – 15 = 2x 2 – 10x + 3x 15 = 2x(x – 5) + 3(x – 5) = (2x + 3)(x – 5) We have, 84 – 2r – 2r 2 = – 2 (r 2 + r – 42) = 2(r 2 + 7r – 6r – 42) = 2[r(r + 7) 6(r + 7)] = 2(6 – r)(r + 7) or 2(6 – r) (7 + r)
Question 24. Factorise: (i) 2x 3 – 3x 2 – 17x + 30 (ii) x 3 6x 2 +11x6 (iii) x 3 + x 2 4x4 (iv) 3x 3 x 2 3x+1 Solution: (i) We have, 2X 3 – 3x 2 – 17x + 30 = 2x 3 – 4x 2 + x 2 – 2x – 15x + 30 = 2x 2 (x – 2) + x(x – 2) – 15(x – 2) = (x – 2) (2x 2 + x – 15) = (x – 2) (2x 2 + 6x – 5x – 15) = (x – 2) [2x(x + 3) – 5(x + 3)] = (x – 2)(x + 3)(2x – 5)
(ii)We have, x 3 – 6x 2 + 11x – 6 = x 3 – x 2 – 5x 2 + 5x + 6x – 6 = x 2 (x – 1) – 5x (x – 1) + 6(x – 1) = (x 1) (x 2 – 5x + 6) = (x – 1) (x 2 – 3x – 2x + 6) = (x – 1) [x(x – 3) – 2(x – 3)] = (x1)(x2)(x3)
(iii) We have, x 3 + x 2 – 4x – 4 = x 2 (x + 1) – 4(x + 1) = (x + 1)(x 2 – 4) = (x + 1) (x – 2) (x + 2)[∴ a 2 – b 2 = (a – b) (a + b)]
(iv) We have, 3x 3 – x 2 – 3x + 1 = 3x 3 – 3x 2 + 2x 2 – 2x – x + 1 = 3x 2 (x – 1) + 2x(x – 1) 1(x – 1) = (x – 1)(3x 2 + 2x – 1) = (x – 1) (3x 2 + 3x – x – 1) = (x – 1) [3x(x + 1) – 1(x + 1)] = (x – 1) (x + 1)(3x 1)
Question 25. Using suitable identity, evaluate the following: (i) 103 3 (ii) 101 × 102 (iii) 999 2 Solution: (i) We have, 103 3 = (100 + 3) 3 = (100) 3 + (3) 3 + 3(100)(3)(100 + 3) [∴ (a + b) 3 = a 3 + b 3 + 3ab(a + b)] = 1000000 + 27 + 900(103) = 1000027 + 92700 = 1092727
(ii) We have, 101 × 102 = (100 + 1) (100 + 2) = (100) 2 + (1 + 2)100 + (1)(2) [∴ (x + a)(x + b) = x 2 + (a + b)x + ab] = 10000 + 300 + 2 = 10302
(iii) We have, (999) 2 = (1000 1) 2 = (1000) 2 + (1) 2 – 2(1000)(1) [∴ (a – b) 2 = a 2 + b 2 – 2ab] = 1000000 + 1 – 2000 = 998001
Question 27. Factorise the following: (i) 9x 2 – 12x + 3 (ii) 9x 2 – 12x + 4 Solution: (i) We have, 9x 2 – 12x + 3 = 3(3x 2 – 4x + 1) = 3(3x 2 – 3x – x + 1) = 3[3x(x – 1) – 1(x – 1)] = 3(3x – 1)(x – 1)
(ii) We have, 9x 2 – 12x + 4 = (3x) 2 – 2 × 3x × 2 + (2) 2 = (3x – 2) 2 [∴ a 2 – 2ab + b 2 = (a – b)²] = (3x – 2) (3x – 2)
Question 28. Expand the following: (i) (4o – b + 2c) 2 (ii) (3o5bc) 2 (ii) (x + 2y – 3z) 2 Solution: We know that (x + y + z) 2 = x 2 + y 2 + z 2 + 2xy + 2yz + 2zx (i) We have, (4a – b + 2c) 2 = (4a) 2 + (b) 2 + (2c) 2 + 2(4a)(b) + 2(b)(2c) + 2(2c)(4a) = 16a 2 + b 2 + 4c 2 – 8ab – 4bc + 16ac
(ii)We have, (3a – 5b – cf = (3a) 2 + (5b² + ( c²) + 2(3a)(5b) + 2(5b)(c) + 2(c)(3a) = 9a 2 + 25b 2 + c 2 – 30ab + 10bc – 6ac
(iii) We have, ( x + 2y – 3z) 2 = ( x) 2 + (2y) 2 + (3z) 2 + 2(x)(2y) + 2(2y)( 3z) + 2( 3z)( x) = x 2 + 4y 2 + 9z 2 – 4xy – 12yz + 6xz
Question 29. Factorise the following: (i) 9x 2 + 4y 2 +16z 2 +12xy16yz24xz (ii) 25x 2 + 16y 2 + 4Z 2 – 40xy +16yz – 20xz (iii) 16x 2 + 4)^ + 9z 2 ^ 6xy – 12yz + 24xz Solution: (i) We have, 9x 2 + 4y 2 + 16z 2 + 12xy – 16yz – 24xz = (3x) 2 + (2y) 2 + (4z) 2 + 2(3x)(2y) + 2(2y)(4z) + 2(4z)(3x) = (3x + 2y – 4z) 2 [∴ a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = (a + b + c) 2 ] = (3x + 2y 4z) (3x + 2y – 4z)
(ii)We have, 25X 2 + 16y 2 + 4z 2 – 40xy + 16yz – 20xz =(5x) 2 + (4y) 2 + (2z) 2 + 2(5x)(4y) + 2(4y)(2z) + 2(2z)(5x) = ( 5x + 4y + 2z) 2 [∴ a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = (a + b + c) 2 ] = ( 5x + 4y + 2z)( 5x + 4y + 2z)
(iii) We have, 16x 2 + 4y 2 + 9z 2 – 16xy – 12yz + 24xz = (4x) 2 + ( 2y) 2 + (3z) 2 + 2(4x)(2y) + 2(2y)(3z) + 2(3z)(4x) = (4x – 2y + 3z) 2 [∴ a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = (a + b + c) 2 ] = (4x – 2y + 3z)(4x 2y + 3z)
Question 30. lf a+b+c=9 and ab + bc + ca = 26, find a 2 + b 2 + c 2 . Solution: We have, a + b + c = 9 ⇒ (a + b + c) 2 = (9) 2 [Squaring on both sides] ⇒ a 2 + b 2 + c 2 + 2ab + 2bc + 2ca = 81 ⇒ a 2 + b 2 + c 2 + 2 (ab + bc + ca) = 81 ⇒ a 2 + b 2 + c 2 + 2(26) = 81 [∴ ab + bc + ca = 26] ⇒ a 2 + b 2 + c 2 = 81 – 52 = 29
Question 34. Factorise: (i) 1 + 64x 3 (ii) a 3 – 2√2b 3 Solution: (i) We have, 1 + 64x 3 = (1) 3 + (4x) 3 = (1 + 4x)[(1) 2 – (1)(4x) + (4x) 2 ] [∴ a 3 + b 3 = (a + b)(a 2 – ab + b 2 )] = (1 + 4x)(1 – 4x + 16x 2 )
(ii) We have, a 3 – 2√2b 3 = (a) 3 – (√2b) 3 = (a – √2b)[a 2 + a( √2b) + (√2b) 2 ] [∴ a 3 – b 3 = (a – b)(a 2 + ab + b 2 )] = (a – √2b)(a 2 + √2ab + 2b 2 )
Question 35. Fi nd (2x – y + 3z) (4x 2 + y 2 + 9z 2 + 2xy + 3yz – 6xz). Solution: We have, (2x – y + 3z) (4x 2 + y 2 + 9z 2 + 2xy + 3yz – 6xz) = 2x(4x 2 + y 2 + 9z 2 + 2xy + 3yz – 6xz) – y(4x 2 + y 2 + 9z 2 + 2xy + 3yz – 6xz) + 3z(4x 2 + y 2 + 9z 2 + 2xy + 3yz – 6xz) = 8x³ + 2xy 2 + 18xz 2 + 4x 2 y + 6xyz – 12x 2 z – 4x 2 y – y 3 – 9yz 2 – 2xy 2 – 3y 2 z + 6xyz + 12x 2 z + 3y 2 z + 27z 3 + 6xyz + 9yz 2 – 18xz 2 = 8X 3 – y 3 + 27z 3 + 18xyz
(ii) We have, (0.2) 3 – (0.3) 3 + (0.1) 3 = (0.2) 3 + ( 0.3) 3 + (0.1) 3 Since, 0.2 – 0.3 + 0.1 = 0, ∴ (0.2)³ + (0.3)³ + (0.1) 3 = 3(0.2) (0.3) (0.1) [ ∴ If a + b + c = 0, then a 3 + b 3 + c 3 = 3abc] = 0.018
Question 38. Without finding the cubes, factorise (x 2y) 3 + (2y – 3z) 3 + (3z – x) 3 . Solution: we see that (x – 2y) +(2y – 3z)+ (3z – x) = 0 Therefore, (x – 2y)³ + (2y – 3z)³ + (3z – x)³ = 3(x – 2y)(2y – 3z)(3z – x). If a + b + c = 0, then a 3 + b 3 + c 3 = 3abc
Question 39. Find the value of (i) x 3 + y 3 – 12xy + 64,when x + y = 4. (ii) x 3 – 8y 3 – 36xy216,when x = 2y + 6. Solution: (i) Since, x + y + 4 = 0, then x 3 + y 3 + (4) 3 = 3xy(4) [∴ If a + b + c = 0, then a 3 + b 3 + c 3 = 3abc] ⇒ x 3 + y 3 + 64 = 12xy ⇒ x 3 + y 3 – 12xy + 64 = 0
(ii) Since, x – 2y – 6 = 0, then x³ + (2y) 3 + (6) 3 = 3x(2y)(6) [∴ If a + b + c = 0, then a 3 + b 3 + c 3 3abc] ⇒ x 3 – 8y 3 – 216 = 36xy ⇒ x 3 – 8y 3 – 36xy – 216 = 0
Question 40. Give possible expression for the length and breadth of the rectangle whose area is given by 4a 2 + 4a – 3. Solution: Given, area of rectangle = (Length) × (Breadth) = 4a 2 + 4a – 3 = 4a 2 + 6a – 2a – 3 = 2a(2a + 3) 1 (2a + 3) = (2a – 1)(2a + 3) Hence, possible length = 2a 1 and breadth = 2a + 3
Exercise 2.4
Question 1. If the polynomials az 3 + 4z 2 + 3z – 4 and z 3 – 4z + o leave the same remainder when divided by z – 3, find the value of a. Solution: Let p 1 (z) = az 3 + 4z 2 + 3z – 4 and p 2 (z) = z 3 – 4z + o When we divide p 1 (z) by z – 3, then we get the remainder p,(3). Now, p 1 (3) = a(3)3 + 4(3)2 + 3(3) – 4 = 27a+ 36+ 94= 27a+ 41 When we divide p 2 (z) by z3 then we get the remainder p 2 (3). Now, p 2 (3) = (3) 3 4(3)+a = 2712 + a = 15+a According to the question, both the remainders are same. p 1 (3) = p 2 (3) 27a + 41 = 15 + a 27a – a = 15 – 41 . 26a = 26 ∴ a = 1
Question 2. The polynomial p{x) = x 4 – 2x 3 + 3x 2 – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also, find the remainder when p(x) is divided by x + 2. Solution: We have, p(x) = x 4 – 2x 3 + 3x 2 – ox + 3a – 7 Since p(x) is divided by x + 1, then remainder is p(1). p(1) = (1) 4 – 2(1) 3 + 3(1) 2 – a(1) + 3a – 7 = 1+ 2 + 3 + o + 3a – 7 = 4a – 1 p( 1) = 19 (Given) ⇒ 4a – 1 = 19 ⇒ 4a = 20 ∴ a = 5 Thus, required polynomial, p(x) = x 4 – 2x 3 + 3x 2 – 5x + 3(5) – 7 = x 4 – 2x 3 + 3x 2 – 5x + 8 Now, this is divided by x + 2, then remainder is p(2). p( 2) = ( 2) 4 – 2( 2) 3 + 3( 2) 2 – 5( 2) + 8 = 16 + 16 + 12 + 10 + 8 = 62
Question 4. Without actual division, prove that 2x 4 – Sx 3 + 2x 2 – x + 2 is divisible by x 2 – 3x + 2. [Hint: Factorise x 2 – 3x + 2] Solution: Let p(x) = 2x 4 – 5x 3 + 2x 2 – x + 2 Now, x 2 3x + 2 = x 2 – 2x – x + 2 = (x2)(x 1) Hence, zeroes of x 2 – 3x + 2 are 1 and 2. ⇒ p(x) is divisible by x 2 – 3x + 2 i.e., divisible by x – 1 and x – 2, if p(1) = 0 and p(2) = 0 Now, p(1) = 2(1) 4 – 5(1) 3 + 2(1) 2 1 + 2 = 2 – 5 + 2 – 1 + 2 = 6 – 6 = 0 and p( 2) = 2(2) 4 – 5(2) 3 + 2(2) 2 2 + 2 = 32 – 40 + 8 = 40 – 40 = 0 Hence, p(x) is divisible by x 2 – 3x + 2.
Question 6. Multiply x 2 + 4y 2 + z 2 + 2xy + xz – 2yz by (z + x2y). Solution: We have, (x 2 + 4y 2 + z 2 + 2xy + xz – 2yz)( z + x – 2y) = x 2 ( z + x – 2y) + 4y 2 ( z + x – 2y) + z 2 ( z + x – 2y) + 2xy( z + x – 2y) + xz( z + x – 2y) – 2yz ( z + x – 2y) = x 2 z + x 3 – 2 x 2 y – 4y 2 z + 4xy 2 – 81y 3 – z 3 + xz 2 – 2yz 2 – 2xyz + 2x 2 y – 4xy 2 – xz 2 + x 2 z – 2xyz + 2yz 2 – 2xyz + 4y 2 z = x 3 – 8y 3 – z 3 – 6xyz
Question 8. If a + b + c = 5 and ab + bc + ca = 10, then prove that a 3 + b 3 + c 3 – 3abc = 25. Solution: We have, a + b + c = 5,ab + bc + ca = 10 Since (a + b + c) 2 = a 2 + b 2 + c 2 + 2(ab + bc + ca), then (5) 2 = a 2 + b 2 + c 2 + 2(10) ⇒ a 2 + b 2 + c 2 = 25 – 20 ⇒ a 2 + b 2 + c 2 = 5 … (i) L.H.S. = a 3 + b 3 + c 3 – 3abc = (a + b + c)(a 2 + b 2 + c 2 – ab – be – ca) = (5) [5 – (ab + be + ca)] [From (i)] = 5(5 10) = 5(5) = – 25 = R.H.S.
Question 9. Prove that (a +b +c) 3 a 3 b 3 – c 3 =3(a +b)(b +c)(c +a). Solution: L.H.S. = [(a + b + c) 3 – a 3 ] – (b 3 + c 3 ) = (a + b + c – a)[(a + b + c) 2 + a 2 + (a + b + c)a] – [(b + c) (b 2 + c 2 – be)] x 3 – y 3 = (x – y)(x 2 + y 2 + xy) and x 3 + y 3 = (x + y)(x 2 + y 2 – xy) = (b + c)[a 2 + b 2 + c 2 + 2 ab + 2 bc + 2 ca + a 2 + a 2 + ab + ac] – (b + c)(b 2 + c 2 – bc) = (b + c)[3a 2 +3ab + 3ac + 3bc] = (b + c)[3(a 2 + ab + ac + bc)] = 3 (b + c)[a(a + b) + c(a + b)] = 3(a + b)(b + c)(c + a) = R.H.S.
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Important Questions with Solutions for CBSE Class 9 Maths Chapter 2  Polynomials
 Class 9 Important Question
 Chapter 2: Polynomials
CBSE Class 9 Maths Chapter2 Important Questions  Free PDF Download
Mathematics is one of the most important disciplines in today's academic curriculum. Maths is concerned with the operation of a certain job statically and quantitatively. It is a complex subject with several themes, formulae, and theories. As a result, students must focus on the topic in order to progress academically. Students must begin with a strong core understanding in order to deal with the difficulties of mathematics in upper grades.
Class 9 is a vital part of a student's academic career as the first board examinations knock at the door in the upcoming year. Therefore they must have a good understanding of crucial subjects like Maths to secure well in the examinations. Here, Vedantu steps in as an efficient guide to the students to provide a push to score well in their examinations through important questions for class 9 Maths Chapter 2 and solutions framed by professionals. You can also Download NCERT Maths Class 9 to help you to revise the complete Syllabus and score more marks in your examinations. Students can also avail of NCERT Solutions for Class 9 Science from our website. Besides, find NCERT Solutions to get more understanding of various subjects. The solutions are uptodate and are sure to help in your academic journey.
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Also, check CBSE Class 9 Maths Important Questions for other chapters:
CBSE Class 9 Maths Important Questions  
Sl.No  Chapter No  Chapter Name 
1  Chapter 1 

2  Chapter 2  Polynomials 
3  Chapter 3 

4  Chapter 4 

5  Chapter 5 

6  Chapter 6 

7  Chapter 7 

8  Chapter 8 

9  Chapter 9 

10  Chapter 10 

11  Chapter 11 

12  Chapter 12 

13  Chapter 13 

14  Chapter 14 

15  Chapter 15 

Important Questions for Class 9 Maths Polynomials
It is the most extensively used subject in science and computer, standing based on logic and reasoning. Therefore, every student must have an indepth learning of the subject to frame a successful career. Students with a weak core of knowledge in Mathematics may face difficulties to deal with the complex formulas and concepts of Maths, which in return serves as a hurdle to score well in the examination. In class 9, algebra is one of the crucial domains of Maths where maximum students fumble.
An efficient learning algebra and its concepts enable the students to dismantle diplomatic Mathematics problems into more straightforward and convenient processes. Therefordents must practice important questions for class 9 maths chapter 2 to master the topic perfectly. Students who find the topic challenging or get stuck with any concept or problems of algebra many refer to the detailed solutions available on the Internet. Quality study material like important questions for class 9 maths polynomials are also availed for the students to access freely in text and PDF formats.
Class 9 Maths Chapter 2 Important Questions
Students are presented with an extensive view of the algebraic concepts and theories in class 9 Maths Ch 2 important questions. To explore different concepts of the chapter and practice a variety of problems, students must have their hands on important polynomials for class 9. Now, let's discuss some of the details about the chapter:
Polynomials
Polynomials can be quoted as an algebraic expression formed using indeterminates or variables and constants or coefficients. This algebraic expression allows such to perform addition, subtraction, multiplication, positive integer exponentiation of variables. The word polynomial is framed from 'poly' meaning 'many' and 'nominal' meaning 'term,' depicting many terms, which means a polynomial contains many terms but not infinite terms.
A polynomial expression comprises variables like x, y,z, coefficients like 1,2, and exponents like 2 in x². The polynomial function is generally depicted by P(x), where x is the variable. For instance,
P(x) = x² + 7x + 15, here x is the variable and 15 is the constant.
Types of Polynomials
Polynomials are categorised into three groups depending upon the number of terms it comprises of. Here are the types of polynomials.
A monomial is a type of polynomial in algebra consisting of a single nonzero term. A polynomial expression consists of one or more terms. Therefore, every term of a polynomial expression is a monomial. Every numeric value such as 6, 12, 151 is a monomial by itself, whereas the variables can x, y, a can also be included in the list of monomials in algebra. Example of a monomial expression – 7x².
Rules for monomial algebraic expressions:
If a monomial is multiplied by a constant, the output will also be a monomial.
If a monomial is multiplied by a monomial, the result will also be a monomial. For instance, if a monomial three is multiplied by 3, the result 8 is also a monomial.
A binomial is a type of polynomial expression comprising of two nonzero terms. Let's see some examples to make it clear,
7x² + 8y is a binomial expression with two variables.
10x⁴ + 9y is also a binomial expression with two variables.
A trinomial is a type of polynomial expression comprising of three nonzero terms. Let's see some examples to make it clear,
5x²+8x+9 is a trinomial expression with one variables x.
a + b+ c is a trinomial expression with three variables.
7x – 6y + 9z is a trinomial expression with three variables.
Students can explore different questions polynomial types through the class 9 Maths chapter 2 important questions.
Polynomial Theorems
Some of the vital theorems of polynomials are as follows:
Remainder Theorem
The polynomial remainder theorem, also quoted as the little Bezout's theorem, implies that if a polynomial P(x) is divided by any linear polynomial depicted by (x – a), the remainder of the operation will be a constant given by P(a), i.e., r = P(a).
Factor Theorem
The factor theorem implies that if P(x) is a polynomial of degree n > 1, and 'a' is a real number, this portrays that:
If P(x) = 0, then (x – a) is the factor of P(x),
If (x – a) is the factor of P(x), P(x) = 0.
Bezout's Theorem
Bezout's Theorem states that if P(x) = 0, then P(x) gets divided by (x – a), with 'r' as the remainder.
Intermediate Value Theorem
The intermediate value theorem states that when a polynomial function transforms from a negative to a positive value, it must cross the xaxis. In other words, the theorem highlights the properties of continuity of a function.
Fundamental Theorem of Algebra
The fundamental theorem of algebra states that each nonconstant single variable that consists of a complex coefficient possess a minimum of one complex root.
Polynomial Equations
A polynomial equation is an algebraic equation comprising of variables with positive integer exponents and constants. A polynomial expression may contain many exponents, and the highest exponent value is termed as the degree of the equation. Let's take an example to make it clear,
ax⁴ + bx² + x + c, is a polynomial expression with degree = 4.
Algebraic identities of polynomials
Identity 1 : (x + z )2 = x² + 2xz + z²
Identity 2 : (y – z) 2 = y² – 2yz + z²
Identity 3 : y² – z² = (y + z) (y – z)
Identity 4 : (x + y) (x + z) = x² + (y + z)x + yz
Important Questions of Polynomials for Class 9
To present the students an insight into the algebraic world, we have highlighted here some of the important questions class 9 Maths chapter 2 , after a proper analysis of sample question papers:
What is a polynomial? Explain with example.
What are the types of polynomial expressions?
Explain the Remainder Theorem with an example.
Prove the Factor theorem of polynomials.
Illustrate Bezout's Theorem, and mention it's importance.
What do you mean by the degree of the polynomial? Explain with examples.
How can we add or subtract polynomials?
Explain the standard form of polynomials.
What do you mean by roots of equations? And how to find them.
Find the roots of polynomial equation, f(x) = x⁴ + 5x² + 7x + 19.
Benefits of Class 9 Maths Ch 2 Important Questions
The students preparing for the boards in the upcoming year must prepare a strong core foundation for developing an indepth logic and understanding of algebra. Therefore they can blend the benefits by practising the class 9th Maths chapter 2 important questions . Here we have listed some of the fruitfulness of class 9 polynomials important questions:
Students can develop deep learning of the topics by exploring different types of questions presented in the important polynomials for class 9.
Vedantu, with an efficient team of topnotch educators, has carefully designed the questions after proper research and analysis of the past year's question papers and sample test papers.
The important questions of ch 2 Maths class 9 are carefully designed under the CBSE board's rules ’ strict guidance.
To perform well in mathematics, academic success is practice; the students must efficiently practice the polynomials class 9 important questions.
To prevent any issues or mistakes in the important questions for class 9 maths polynomials , expert teachers have reviewed and analysed the papers.
Mathematics is the foundation for logic and reasoning. As a result, in order to grasp the topic's various subjects, students must work with insufficient fundamental Mathematics comprehension and study important polynomial questions for class 9. Students must have a good comprehension of the crucial questions for class 9 mathematics chapter 2 in order to begin a career in science and technology.
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FAQs on Important Questions with Solutions for CBSE Class 9 Maths Chapter 2  Polynomials
1. Where can I find some important questions for Class 9 Maths Chapter 2?
Ans: Crucial Class 9 Maths Chapter 2 questions titled Polynomials assist students in preparing for the Class 9 Mathematics Test. This will give you a general sense of the types of questions you could encounter in the test and from which chapter. Understanding what to study in a subject makes learning easier and faster since it requires less time. Hence, while preparing for Class 9 Maths Chapter 2 named Polynomials in CBSE Class 9 Mathematics, students in Class 9th are encouraged to answer these key questions.
2. Does Vedantu provide solutions to Class 9 Maths Chapter 2 of NCERT textbook?
Ans: Free PDF download of Important Questions with solutions for CBSE Class 9 Maths Chapter 2 named Polynomials are prepared by Vedantu’s inhouse expert Mathematics teachers from the latest edition of NCERT textbooks. You can register online for Maths tuition on Vedantu if you are eager to score more marks in the final examination. You can also Download NCERT Maths Class 9 Solutions in order to help yourself revise the complete syllabus and score more marks in your examinations. All the Class 9th students can also avail of NCERT Solutions for Science from Vedantu website and mobile application very easily which help them to prepare for the exam along with the important questions.
3. Can I download the solutions to Important Questions with solutions for CBSE Class 9 Maths Chapter 2 offered by Vedantu?
Ans: Yes, you can download the solutions to Important Questions with solutions for CBSE Class 9 Maths Chapter 2 offered by Vedantu. These are available on Vedantu’s official website and mobile app at free of cost in PDF format. In that case, you can download the Vedantu app from the Google play store to avail these Important Questions with solutions for CBSE Class 9 Maths Chapter 2 offered by Vedantu.
4. What is taught in Class 9 Maths Chapter 2 of CBSE curriculum?
Ans: Polynomials is the second chapter of Class 9 Maths. Polynomials are introduced and discussed in detail here. The chapter discusses the Polynomials and their applications. The introduction of the chapter includes whole numbers, integers, and rational numbers.
The chapter begins with the introduction of Polynomials in section 2.1 followed by two other very important topics explained in section 2.2 and 2.3.
Section 2.1  Polynomials in one Variable – This topic discusses the Linear, Quadratic and Cubic Polynomial.
Section 2.2  Zeros of a Polynomial – This chapter explains that a zero of a polynomial need not be zero and can have more than one zero.
Section 2.3  Real Numbers and Their Decimal Expansions – Here you will study the decimal expansions of real numbers and understand if it can help in distinguishing between rationals and irrationals.
5. Can I print the MCQ Questions for Chapter 2 of Class 9 Maths with answers?
Ans: Yes, the MCQ Questions and Answers for Chapter 2 of Class 9 Maths are in a downloadable PDF format and can be printed easily. These are available on Vedantu's website and app and you can download them according to your comfort and timing and can print the MCQs for future reference. Studying these important questions will ensure you get a good score in the final exam for Class 9 Maths.
6. Are these free or is there any charge for the MCQ Questions for Chapter 2 of Class 9 Maths with answers?
Ans: Yes, the MCQ Questions and Answers for Chapter 2 of Class 9 Maths are absolutely free and do not carry any hidden charge or cost. You can also download these from the Vedantu app. You can download these anytime according to your comfort so that you can study as and when required. These important questions have been carefully selected by the experts at Vedantu and are guaranteed to help you to score the best.
7. How Many questions are there in NCERT Solutions of Chapter 2 of Class 9 Maths?
Ans: The question breakup of the exercises covered in NCERT Solutions of Chapter 2 of Class 9 Maths are as follows:
Exercise 2.1 includes five questions
Exercise 2.2 includes four questions
Exercise 2.3 includes three questions
8. What are the Important Topics covered in NCERT Solutions of Chapter 2 of Class 9 Maths?
Ans: The major topics covered in NCERT Solutions for Chapter 2 of Class 9 Maths are as follows:
Polynomials in One Variable
Zeros of a Polynomial
Factorisation of Polynomials
Algebraic Identities
9. Why should I opt for NCERT Solutions of Chapter 2 of Class 9 Maths?
Ans: NCERT Solutions for Class 9 Mathematics Chapter 2 will certainly give you an advantage. The NCERT Solutions are created by experts in the field at Vedantu to provide students with effective solutions to problems while explaining their concepts so that they are never stuck with the same issue again in the future. This will ensure that you get the greatest grades and fully comprehend complicated ideas. As a result, you must without a doubt select NCERT Solutions for Chapter 2 of Class 9 Mathematics.
CBSE Class 9 Maths Important Questions
Cbse study materials.
CBSE Class 9 Maths 30 Most Important Case Study Questions with Answers
Cbse class 9 maths 30 most important case study questions with answers download here free in pdf format..
CBSE Class 9 Maths exam 2023 will have a set of questions based on case studies in the form of MCQs. CBSE Class 9 Maths Question Bank on Case Studies given in this webpage can be very helpful in understanding the new format of questions.
Each question has five subquestions, each followed by four options and one correct answer. Candidates can easily download these questions in PDF format and refer to them for exam preparation 2023.
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 Circles Class 9 Case Study Questions Maths Chapter 9
Last Updated on September 8, 2024 by XAM CONTENT
Hello students, we are providing case study questions for class 9 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 9 maths. In this article, you will find case study questions for CBSE Class 9 Maths Chapter 9 Circles. It is a part of Case Study Questions for CBSE Class 9 Maths Series.
Circles  
Case Study Questions  
Competency Based Questions  
CBSE  
9  
Maths  
Class 9 Studying Students  
Yes  
Mentioned  
Case Study Questions on Circles
Government of India is working regularly for the growth of handicapped persons. For these three STD booths situated at point P, Q and R are as shown in the figure, which are operated by handicapped persons. These three booths are equidistant from each other as shown in the figure.
On the basis of the above information, solve the following questions:
Q. 1. Which type of ΔPQR in the given figure?
Q. 2. Measure angle ∠QOR.
Q. 3. Find the value of ∠OQR.
Q. 4. Is it true that points P, Q and R lie on the circle?
1. Given $P, Q$ and $R$ are equidistant. It means their distances are equal.
Note: In an equilateral triangle, length of all three sides are equal. So, $\triangle P Q R$ is an equilateral triangle.
2. Since, $\triangle P Q R$ is an equilateral triangle.
$$ \therefore \angle \mathrm{PQR}=\angle \mathrm{PRQ}=\angle \mathrm{QPR}=60^{\circ} $$
The angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.
$$ \therefore \angle Q O R=2 \angle Q P R=2 \times 60^{\circ}=120^{\circ} $$
3. In $\triangle O Q R$,
$$ O Q=O R[\text { Radii of a circle }) $$
$\Rightarrow \angle \mathrm{ORQ}=\angle \mathrm{OQR}$ [Angles opposite to equal sides of a triangle are equal] Using angle sum property of a triangle,
$$ \begin{aligned} & \angle \mathrm{OQR}+\angle \mathrm{ORQ}+\angle \mathrm{QOR}=180^{\circ} \\ & \Rightarrow \angle \mathrm{OQR}+\angle \mathrm{OQR}+120^{\circ}=180^{\circ} \end{aligned} $$
$$ \begin{gathered} \Rightarrow 2 \angle O Q R=60^{\circ} \\ \therefore \angle O Q R=30^{\circ} \end{gathered} $$
4. Yes, it is true that points $P, Q$ and $R$ lie on the circle.
Understanding Circles
Circle: A collection of all points in a plane which are at a constant distance from a fixed point. The fixed point is called the centre of the circle and the constant distance is called the radius.
In figure, O is the centre, OA is the radius and AB is the diameter of the circle.
Chord: A line segment joining any two points on the circle. In figure, AC is the chord.
Diameter: Longest chord of the circle that passes through the centre of the circle.
Circumference: Length of the boundary of a circle.
Arc: Any part of the circumference of a circle.
In figure, PRQ is minor arc represented as PRQ and PSQ is major arc represented as PSQ.
Semicircle: Parts of a circle that are divided by its diameter.
Segment: The region between a chord and either of its arcs (major or minor). The segment formed with a minor arc is called minor segment and that formed with a major arc is called major segment.
Sector: The region enclosed by an arc and the two radii joining the centre to the end points of the arc.
The sector corresponding to minor arc is called minor sector i.e., AOB and that corresponding to major arc is called major sector. i.e., AOBCA
 Heron’s Formula Class 9 Case Study Questions Maths Chapter 10
 Quadrilaterals Class 9 Case Study Questions Maths Chapter 8
 Triangles Class 9 Case Study Questions Maths Chapter 7
 Lines and Angles Class 9 Case Study Questions Maths Chapter 6
 Introduction to Euclid’s Geometry Class 9 Case Study Questions Maths Chapter 5
 Linear Equations in Two Variables Class 9 Case Study Questions Maths Chapter 4
 Coordinate Geometry Class 9 Case Study Questions Maths Chapter 3
Polynomials Class 9 Case Study Questions Maths Chapter 2
Number systems class 9 case study questions maths chapter 1, topics from which case study questions may be asked.
 Equal chords of a circle subtend equal angles at the center and (motivate) its converse.
 The perpendicular from the center of a circle to a chord bisects the chord and onversely, the line drawn through the center of a circle to bisect a chord is perpendicular to the chord.
 Equal chords of a circle (or of congruent circles) are equidistant from the center (or their respective centers) and conversely.
 The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
 Angles in the same segment of a circle are equal.
 If a line segment joining two points subtends equal angle at two other points lying on the same side of the line containing the segment, the four points lie on a circle.
 The sum of either of the pair of the opposite angles of a cyclic quadrilateral is 180° and its converse.
Circles having same centre are called concentric circles.
Case study questions from the above given topic may be asked.
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Frequently Asked Questions (FAQs) on Circles Case Study
Q1: what is the definition of a circle in class 9 maths.
A1: A circle is defined as the collection of all points in a plane that are at a fixed distance (radius) from a fixed point called the center. In Class 9, the concept of a circle is used to study properties related to chords, tangents, arcs, and sectors.
Q2: What is the difference between a chord and a diameter?
A2: A chord is a line segment joining two points on the circumference of a circle, whereas the diameter is a special chord that passes through the center of the circle. The diameter is the longest chord of the circle and is equal to twice the radius.
Q3: What are tangents and how are they important in this chapter?
A3: A tangent to a circle is a line that touches the circle at exactly one point. Tangents are important in this chapter as you learn properties like “a tangent to a circle is perpendicular to the radius at the point of contact.” You also explore problems involving two tangents drawn from an external point.
Q4: What are the main properties of circles discussed in Class 9 Maths?
A4: The main properties include: (i) The radius is perpendicular to the tangent at the point of contact. (ii) Equal chords are equidistant from the center. (iii) The angle subtended by a chord at the center of the circle is twice the angle subtended by the same chord at any other point on the circumference.
Q5: What are cyclic quadrilaterals, and how are they related to circles?
A5: A cyclic quadrilateral is a foursided figure where all its vertices lie on the circumference of a circle. The sum of the opposite angles in a cyclic quadrilateral is always 180°. This property is crucial in solving various geometric problems involving circles.
Q6: How can we prove that the angle subtended by a diameter on the circumference is 90 degrees?
A6: This is a key result derived from the properties of circles. Using the theorem “the angle subtended by a chord at the center is twice the angle subtended at any other point on the circumference,” and since a diameter subtends a straight angle (180°) at the center, it subtends a right angle (90°) on the circumference.
Q7: How many tangents can be drawn from a point outside a circle?
A7: From any point outside a circle, exactly two tangents can be drawn. These tangents have equal lengths from the external point to their points of contact on the circle.
Q8: What are the most important theorems in Chapter 9 of Class 9 Maths?
A8: The two key theorems in this chapter are: (i) The perpendicular from the center of a circle to a chord bisects the chord. (ii) The length of tangents drawn from an external point to a circle are equal.
Q9: Are there any online resources or tools available for practicing Circles case study questions?
A9: We provide case study questions for CBSE Class 9 Maths on our website. Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit Physics Gurukul website. they are having a large collection of case study questions for all classes.
 CBSE Class 10
CBSE Class 10 Maths CompetencyBased Questions With Answer Key 202425: Chapter 6 Triangles Free PDF Download
Cbse class 10 maths chapter 6 practice questions 2025: students check and download the cbse class 10 chapter 6 triangles competencyfocused practice questions along with answer key for 202425. .
CBSE 202425 Competency Based Questions With Answers: The Central Board of Secondary Education (CBSE) has designed and released the competency‐focused practice questions for students of classes 10 and 12. Students can practice questions ranging from multiple choice questions to case studies. CBSE has made available the practice questions on its website, cbse.academic.nic.in. This article covers the CBSE Class 10 Maths Chapter 6 Triangles competencyfocused questions.
These assessments contain high quality questions and concepts that are important for learning. When students will solve the competencybased questions, they will engage in learning by doing. The questions are designed to test the understanding of students and remove any gaps that may exist in the learning process. Competencybased practice questions follow the principle of learning with understanding.
CBSE Class 10 Maths Chapter 6 Competency BasedQuestions
Give an example each for when two rectangles are: 
A painted cuboid, Solid P, is cut along a plane to get two identical solids. One of the identical solids, Solid Q, is shown below. 
Shown below is an isosceles rightangled ΔPQR. The area of ΔPQR is 18 cm2. 
Shown below is a figure. ΔPQR is a rightangled triangle. There are 3 semicircles with diameters as sides of ΔPQR. All length measurements are in cm. 
Shown below is a figure with two rectangles. The ratio of UV:VW = QR:RS = 3:4. Area of TUVW is 36 cm2. 
i) Two chords of a circle, PQ and MN intersect at a point T. Show that PT × TQ = MT × TN. Draw a figure. 
In the figure below, P, Q and R are collinear. P and Q are centres of the two circles. P lies on the circumference of the circle with centre Q. R is 10 cm from Q and 15 cm from P. Both circles have 2 common tangents from point R. 
Tanmay had hit a numbered ball into pocket P. The path followed by the cue ball and ball 8 is shown below. The cue ball's initial distance from the edges of the table is marked in the figure. 
Shown below is the path when Tanmay hits a ball number 1 in pocket S. The distance CT is 50 cm and TS is 120 cm. The cue ball hits the side PQ at an angle of 45°. 
Tips to Prepare for CBSE Class 10 Mathematics Examination
1. Students must first familiarize themselves with the syllabus so as to understand which topics carry more weightage. Ensure that you prepare according to the latest syllabus only.
2. Mathematics is a subject that demands regular practice. It is important that students practice as much as they can. Practice the questions given at the end of each chapter. Note the points where you get stuck during solving questions.
3. Students should prepare an effective study plan and give time to all topics on the basis of their strengths and weaknesses. Don’t let your bias towards your favourite chapter affect the timetable. The schedule should also have time for revision.
4. Students should solve the problems from CBSE Class 10 subject textbook to clear the concepts and strengthen understanding. While solving problems, students should make it a habit to show rough work clearly which will help them in the examination as well.
CBSE Video Courses for Class 10 Students
Class 10 students can study effectively for the exams with the help of video courses prepared by the subject matter experts. These video courses will explain the concepts in a simple and interactive manner which will help learners to understand clearly.
CBSE Class 10 Video Courses
Also, check
CBSE Class 10 Syllabus 202425: Download Latest and Revised FREE PDFs
NCERT Books for Class 10 All Subjects PDF (202425)
NCERT Solutions for Class 10 (20242025) All Subjects & Chapters: Download in PDF
CBSE Class 10 Maths Mind Maps for Quick Revision
CBSE Class 10 Maths MCQs: Important for 20242025 Exams, Download ChapterWise PDF
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Here, we have provided casebased/passagebased questions for Class 9 Maths Chapter 2 Polynomials. Case Study/Passage Based Questions. Ankur and Ranjan start a new business together. The amount invested by both partners together is given by the polynomial p (x) = 4x 2 + 12x + 5, which is the product of their individual shares.
Find the total length of the glass portion of the door (in inches) is represented in terms of x. Q. 2. Find the total width of the glass portion of the door (in inches). Q. 3. Write the polynomial representation of the area top half part of the door. Q. 4. Find the zeroes of the polynomial representing the area.
Download Class 9 Maths Case Study Questions to prepare for the upcoming CBSE Class 9 Exams 202324. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 9 so that they can score 100% in Exams. Case study questions play a pivotal role in enhancing students' problemsolving skills.
CBSE Class 9th Maths 2023 : 30 Most Important Case ...
CBSE Case Study Questions for Class 9 Maths
Case Studies In Class 9 Mathematics. The Central Board of Secondary Education (CBSE) has included case study based questions in the Class 9 Mathematics paper in current session. According to new pattern CBSE Class 9 Mathematics students will have to solve case based questions.
The above Case studies for Class 9 Mathematics will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Maths Case Studies have been developed by experienced teachers of schools.studyrate.in for benefit of Class 10 students. Class 9 Science Case Study Questions.
Important Questions Class 9 Maths Chapter 2  Polynomials
Important Questions for CBSE Class 9 Maths Chapter 2 ...
Case Study Questions Class 9 Maths Chapter 2. Case Study/PassageBased Questions. Case Study 1. Ankur and Ranjan start a new business together. The amount invested by both partners together is given by the polynomial p (x) = 4x 2 + 12x + 5, which is the product of their individual shares. Coefficient of x2 in the given polynomial is.
Polynomials Case Study Questions (CSQ's) Select the number of questions for the test: Keep paper and pencil ready but keep your books away. You can move between questions and answer them in any order you like. These tests are unlimited in nature…take as many as you like. You will be able to view the solutions only after you end the test.
CBSE Class 9 Maths Chapter 2  Polynomials Important ...
CBSE Class 9 Maths Question Bank on Case Studies given in this webpage can be very helpful in understanding the new format of questions. Each question has five subquestions, each followed by four options and one correct answer. Candidates can easily download these questions in PDF format and refer to them for exam preparation 2023.
Chapter 2 Polynomials Class 9 Maths NCERT Solutions PDF download is very useful in understanding the basic concepts embedded in the chapter. It is very essential to solve every question before moving further to any other supplementary books.
1) x 22x+1 is a polynomial in: a. One Variable b. Two Variables c. Three variable d. None of the above. Answer/Explanation. Answer: (a) Explanation: x 22x+1 can be written as x 22x 1 +1x 0.Hence, we can see that x is the only variable having powers as whole numbers: 2,1 and 0.
Get CBSE Class 9 Maths Extra Questions and Answers for Chapter 2 Polynomials. All the extra questions are based on the NCERT Book. Practice with theses important questions to score good marks in ...
Reading Time: 10 minutes Last Updated on September 8, 2024 by XAM CONTENT. Hello students, we are providing case study questions for class 9 maths. Case study questions are the new question format that is introduced in CBSE board.
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CBSE Class 10 Maths CompetencyBased Questions With Answers 202425: CBSE Class 10 students download competencyfocused practice questions and answers for Maths chapter 6 Triangles. Download free ...
 Circles Class 9 Case Study Questions Maths Chapter 9
Last Updated on September 8, 2024 by XAM CONTENT
Hello students, we are providing case study questions for class 9 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 9 maths. In this article, you will find case study questions for CBSE Class 9 Maths Chapter 9 Circles. It is a part of Case Study Questions for CBSE Class 9 Maths Series.
Circles  
Case Study Questions  
Competency Based Questions  
CBSE  
9  
Maths  
Class 9 Studying Students  
Yes  
Mentioned  
Table of Contents
Case Study Questions on Circles
Government of India is working regularly for the growth of handicapped persons. For these three STD booths situated at point P, Q and R are as shown in the figure, which are operated by handicapped persons. These three booths are equidistant from each other as shown in the figure.
On the basis of the above information, solve the following questions:
Q. 1. Which type of ΔPQR in the given figure?
Q. 2. Measure angle ∠QOR.
Q. 3. Find the value of ∠OQR.
Q. 4. Is it true that points P, Q and R lie on the circle?
1. Given $P, Q$ and $R$ are equidistant. It means their distances are equal.
Note: In an equilateral triangle, length of all three sides are equal. So, $\triangle P Q R$ is an equilateral triangle.
2. Since, $\triangle P Q R$ is an equilateral triangle.
$$ \therefore \angle \mathrm{PQR}=\angle \mathrm{PRQ}=\angle \mathrm{QPR}=60^{\circ} $$
The angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.
$$ \therefore \angle Q O R=2 \angle Q P R=2 \times 60^{\circ}=120^{\circ} $$
3. In $\triangle O Q R$,
$$ O Q=O R[\text { Radii of a circle }) $$
$\Rightarrow \angle \mathrm{ORQ}=\angle \mathrm{OQR}$ [Angles opposite to equal sides of a triangle are equal] Using angle sum property of a triangle,
$$ \begin{aligned} & \angle \mathrm{OQR}+\angle \mathrm{ORQ}+\angle \mathrm{QOR}=180^{\circ} \\ & \Rightarrow \angle \mathrm{OQR}+\angle \mathrm{OQR}+120^{\circ}=180^{\circ} \end{aligned} $$
$$ \begin{gathered} \Rightarrow 2 \angle O Q R=60^{\circ} \\ \therefore \angle O Q R=30^{\circ} \end{gathered} $$
4. Yes, it is true that points $P, Q$ and $R$ lie on the circle.
Understanding Circles
Circle: A collection of all points in a plane which are at a constant distance from a fixed point. The fixed point is called the centre of the circle and the constant distance is called the radius.
In figure, O is the centre, OA is the radius and AB is the diameter of the circle.
Chord: A line segment joining any two points on the circle. In figure, AC is the chord.
Diameter: Longest chord of the circle that passes through the centre of the circle.
Circumference: Length of the boundary of a circle.
Arc: Any part of the circumference of a circle.
In figure, PRQ is minor arc represented as PRQ and PSQ is major arc represented as PSQ.
Semicircle: Parts of a circle that are divided by its diameter.
Segment: The region between a chord and either of its arcs (major or minor). The segment formed with a minor arc is called minor segment and that formed with a major arc is called major segment.
Sector: The region enclosed by an arc and the two radii joining the centre to the end points of the arc.
The sector corresponding to minor arc is called minor sector i.e., AOB and that corresponding to major arc is called major sector. i.e., AOBCA
 Heron’s Formula Class 9 Case Study Questions Maths Chapter 10
 Quadrilaterals Class 9 Case Study Questions Maths Chapter 8
 Triangles Class 9 Case Study Questions Maths Chapter 7
 Lines and Angles Class 9 Case Study Questions Maths Chapter 6
 Introduction to Euclid’s Geometry Class 9 Case Study Questions Maths Chapter 5
 Linear Equations in Two Variables Class 9 Case Study Questions Maths Chapter 4
 Coordinate Geometry Class 9 Case Study Questions Maths Chapter 3
Polynomials Class 9 Case Study Questions Maths Chapter 2
Number systems class 9 case study questions maths chapter 1, topics from which case study questions may be asked.
 Equal chords of a circle subtend equal angles at the center and (motivate) its converse.
 The perpendicular from the center of a circle to a chord bisects the chord and onversely, the line drawn through the center of a circle to bisect a chord is perpendicular to the chord.
 Equal chords of a circle (or of congruent circles) are equidistant from the center (or their respective centers) and conversely.
 The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
 Angles in the same segment of a circle are equal.
 If a line segment joining two points subtends equal angle at two other points lying on the same side of the line containing the segment, the four points lie on a circle.
 The sum of either of the pair of the opposite angles of a cyclic quadrilateral is 180° and its converse.
Circles having same centre are called concentric circles.
Case study questions from the above given topic may be asked.
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Frequently Asked Questions (FAQs) on Circles Case Study
Q1: what is the definition of a circle in class 9 maths.
A1: A circle is defined as the collection of all points in a plane that are at a fixed distance (radius) from a fixed point called the center. In Class 9, the concept of a circle is used to study properties related to chords, tangents, arcs, and sectors.
Q2: What is the difference between a chord and a diameter?
A2: A chord is a line segment joining two points on the circumference of a circle, whereas the diameter is a special chord that passes through the center of the circle. The diameter is the longest chord of the circle and is equal to twice the radius.
Q3: What are tangents and how are they important in this chapter?
A3: A tangent to a circle is a line that touches the circle at exactly one point. Tangents are important in this chapter as you learn properties like “a tangent to a circle is perpendicular to the radius at the point of contact.” You also explore problems involving two tangents drawn from an external point.
Q4: What are the main properties of circles discussed in Class 9 Maths?
A4: The main properties include: (i) The radius is perpendicular to the tangent at the point of contact. (ii) Equal chords are equidistant from the center. (iii) The angle subtended by a chord at the center of the circle is twice the angle subtended by the same chord at any other point on the circumference.
Q5: What are cyclic quadrilaterals, and how are they related to circles?
A5: A cyclic quadrilateral is a foursided figure where all its vertices lie on the circumference of a circle. The sum of the opposite angles in a cyclic quadrilateral is always 180°. This property is crucial in solving various geometric problems involving circles.
Q6: How can we prove that the angle subtended by a diameter on the circumference is 90 degrees?
A6: This is a key result derived from the properties of circles. Using the theorem “the angle subtended by a chord at the center is twice the angle subtended at any other point on the circumference,” and since a diameter subtends a straight angle (180°) at the center, it subtends a right angle (90°) on the circumference.
Q7: How many tangents can be drawn from a point outside a circle?
A7: From any point outside a circle, exactly two tangents can be drawn. These tangents have equal lengths from the external point to their points of contact on the circle.
Q8: What are the most important theorems in Chapter 9 of Class 9 Maths?
A8: The two key theorems in this chapter are: (i) The perpendicular from the center of a circle to a chord bisects the chord. (ii) The length of tangents drawn from an external point to a circle are equal.
Q9: Are there any online resources or tools available for practicing Circles case study questions?
A9: We provide case study questions for CBSE Class 9 Maths on our website. Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit Physics Gurukul website. they are having a large collection of case study questions for all classes.
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Here, we have provided casebased/passagebased questions for Class 9 Maths Chapter 2 Polynomials. Case Study/Passage Based Questions. Ankur and Ranjan start a new business together. The amount invested by both partners together is given by the polynomial p (x) = 4x 2 + 12x + 5, which is the product of their individual shares.
Find the total length of the glass portion of the door (in inches) is represented in terms of x. Q. 2. Find the total width of the glass portion of the door (in inches). Q. 3. Write the polynomial representation of the area top half part of the door. Q. 4. Find the zeroes of the polynomial representing the area.
Download Class 9 Maths Case Study Questions to prepare for the upcoming CBSE Class 9 Exams 202324. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 9 so that they can score 100% in Exams. Case study questions play a pivotal role in enhancing students' problemsolving skills.
CBSE Class 9 Maths Question Bank on Case Studies given in this article can be very helpful in understanding the new format of questions. Each question has five subquestions, each followed by four options and one correct answer. Students can easily download these questions in PDF format and refer to them for exam preparation. Case Study Questions.
Case Study Questions Question 1: On one day, principal of a particular school visited the classroom. Class teacher was teaching the concept of polynomial to students. He was very much impressed by her way of teaching. To check, whether the students also understand the concept taught by her or not, he asked variousquestions to students. … Continue reading Case Study Questions for Class 9 ...
In this post I have provided CBSE Class 9 Maths Case Study Based Questions With Solution. ... each followed by four options and one correct answer. All CBSE Class 9th Maths Students can easily download these questions in PDF form with the help of given download Links and refer for exam preparation. ... Casebased Questions  9: PDF Link: Case ...
Class 9 Mathematics Case study question 2. Read the Source/Text given below and answer any four questions: Maths teacher draws a straight line AB shown on the blackboard as per the following figure. Now he told Raju to draw another line CD as in the figure. The teacher told Ajay to mark ∠ AOD as 2z.
CBSE Class 9 Maths Board Exam will have a set of questions based on case studies in the form of MCQs.The CBSE Class 9 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends. If you want to want to prepare all the tough, tricky & difficult questions for your upcoming exams, this is ...
Given below are a few of the questions and answers from our question bank of Important Questions Class 9 Mathematics Chapter 2: Question 1: Calculate the value of 9x² + 4y² if xy = 6 and 3x + 2y = 12. Answer 1: Consider the equation 3x + 2y = 12. Now, square both sides: (3x + 2y)² = 12². => 9x² + 12xy + 4y² = 144. =>9x² + 4y² = 144 ...
78 Views. 26 Downloads. Class 9 Polynomials Case Study Questions  Podar International School Class 9 Mathematics Questions with answers in PDF format for free Class 9 Maths Question Papers with answers Class 9 Podar International School Mathematics CBSE 2024. Download Paper.
Here are some tips to effectively answer case study questions in Class 9 Maths: 1. Read the case study carefully and understand the given information. 2. Identify the mathematical concepts or formulas that are relevant to the case study. 3.
These questions will help the 9th class students to get acquainted with a wide variety of questions and develop the confidence to solve polynomial questions more efficiently. 1. Give an example of a monomial and a binomial having degrees of 82 and 99, respectively. Solution: An example of a monomial having a degree of 82 = x 82.
CBSE Class 9 Maths Question Bank on Case Studies given in this webpage can be very helpful in understanding the new format of questions. Each question has five subquestions, each followed by four options and one correct answer. Candidates can easily download these questions in PDF format and refer to them for exam preparation 2023.
Ans: Crucial Class 9 Maths Chapter 2 questions titled Polynomials assist students in preparing for the Class 9 Mathematics Test. This will give you a general sense of the types of questions you could encounter in the test and from which chapter. Understanding what to study in a subject makes learning easier and faster since it requires less time.
9. Factorise: (a  b) 3 + (b  c) 3 + (c  a) 3. Answer: 3 (a 2 (c−b) + b 2 (a−c) + c 2 (b−a)) 10. If p (x) is a polynomial of degree n > 1 and a is any real number, then (i) x  a ...
Polynomials Case Study Questions (CSQ's) Select the number of questions for the test: Keep paper and pencil ready but keep your books away. You can move between questions and answer them in any order you like. These tests are unlimited in nature…take as many as you like. You will be able to view the solutions only after you end the test.
Free PDF download of NCERT Exemplar for Class 9 Maths Chapter 2  Polynomials solved by expert Maths teachers on Vedantu as per NCERT (CBSE) Book guidelines. All Chapter 2  Polynomials exercise questions with solutions to help you to revise the complete syllabus and score more marks in your examinations. Download free Class 9 Maths to amp up ...
Also, these class 9 maths NCERT solutions Chapter 2 define the algebraic identities, which help in factorizing the algebraic equations. Total Questions: Class 9 Maths Chapter 2 Polynomials consists of a total of 45 questions, of which 31 are easy, 9 are moderate, and 5 are long answer type questions. Cuemath is one of the world's leading math ...
Case Study Questions Class 9 Maths Chapter 2. Case Study/PassageBased Questions. Case Study 1. Ankur and Ranjan start a new business together. The amount invested by both partners together is given by the polynomial p (x) = 4x 2 + 12x + 5, which is the product of their individual shares. Coefficient of x2 in the given polynomial is.
Jul 21, 2022, 16:45 IST. Find below Important questions of class 9 maths of chapter Polynomials prepared by the academic team of Physics Wallah. All important questions of chapter Polynomials class 9 maths are uploaded in pdf form with detail stepbystep solutions. Do solve NCERT questions and for reference use NCERT solutions for class 9 ...
Chapter 2 Polynomials Class 9 Maths NCERT Solutions PDF download is very useful in understanding the basic concepts embedded in the chapter. It is very essential to solve every question before moving further to any other supplementary books.
2.1 Introduction. NCERT Solutions for class 9 chapter 2 polynomials gives you a thorough understanding of the topic. Moreover, our experts give proper attention to the Remainder Theorem and Factor Theorem uses and methods. These subheadings allow the student to grasp the topics in a much easier way.
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Reading Time: 10 minutes Last Updated on September 8, 2024 by XAM CONTENT. Hello students, we are providing case study questions for class 9 maths. Case study questions are the new question format that is introduced in CBSE board.