case study questions class 9 maths chapter 2

Class 9 Maths Case Study Questions of Chapter 2 Polynomials PDF Download

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Case study Questions in Class 9 Mathematics Chapter 2  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  Class 9 Maths Case Study Questions  Chapter 2 Polynomials

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These case study questions challenge students to apply their knowledge of polynomials in real-life scenarios, enhancing their problem-solving abilities. This article provides for the Class 9 Maths Case Study Questions of Chapter 2: Polynomials , enabling students to practice and excel in their examinations.

Polynomials Case Study Questions With Answers

Here, we have provided case-based/passage-based 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.

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)

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/Passage-Based Questions

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

The Class 9 Maths Case Study Questions of Chapter 2: Polynomials serve as a valuable resource for students seeking to enhance their understanding of polynomial concepts and problem-solving skills. By practicing these case studies, students can strengthen their grasp of polynomials and their applications in real-life scenarios. Embrace the opportunity to engage with practical problems and excel in your mathematical journey.

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I didnt understand last question

The number of zeroes of the polynomial x2 + 4x + 2 is The answer is too easy, i.e. 2

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Case Study Questions for Class 9 Maths Chapter 2 Polynomials

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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 9th Maths 2023 : 30 Most Important Case Study Questions with Answers; Download PDF

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CBSE Class 9 Maths exam 2022-23 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.

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CBSE Class 9 Mathematics Case Study Questions

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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 problem-solving and critical-thinking 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 real-life 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 real-life 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 real-world 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)

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 problem-solving 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)

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14 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) (a-b)(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.

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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. Chapter-wise 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.

Cuemath is one of the world's leading math learning platforms that offers LIVE 1-to-1 online math classes for grades K-12 . Our mission is to transform the way children learn math, to help them excel in school and competitive exams. Our expert tutors conduct 2 or more live classes per week, at a pace that matches the child's learning needs.

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

Video solutions for class 9 maths ncert chapter 2, 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 long-form 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.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials are provided here. Our NCERT Maths solutions contain all the questions of the NCERT textbook that are solved and explained beautifully. Here you will get complete NCERT Solutions for Class 9 Maths Chapter 2 all exercises Exercise in one place. These solutions are prepared by the subject experts and as per the latest NCERT syllabus and guidelines. CBSE Class 9 Students who wish to score good marks in the maths exam must practice these questions regularly.

Class 9 Maths Chapter 2 Polynomials NCERT Solutions

Below we have provided the solutions of each exercise of the chapter. Go through the links to access the solutions of exercises you want. You should also check out our NCERT Class 9 Solutions for other subjects to score good marks in the exams.

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.1

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.5

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials – Topic Discussion

Below we have listed the topics that have been discussed in this chapter. As this is one of the important topics in maths, It comes under the unit – Algebra which has a weightage of 20 marks in class 9 maths board exams.

• Polynomials in one Variable – Discussion of Linear, Quadratic and Cubic Polynomial.
• Zeroes of a Polynomial
• Real Numbers and their Decimal Expansions
• Representing Real Numbers on the Number Line Operations on Real Numbers
• Laws of Exponents for Real Numbers.

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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:

How can i download chapter 2 polynomials class 9 ncert solutions pdf , what do you mean by algebraic expressions, what is the coefficient of a zero polynomial, what is biquadratic polynomial, contact form.

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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

• Checkout: Class 9 Science Case Study Questions
• Checkout: Class 9 Maths Case Study Questions

Polynomials Case Study Questions With Answers

Case study questions class 9 maths chapter 2.

Case Study/Passage-Based 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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• NCERT Solutions for Class 9 Maths Chapter 2: Polynomials - Exercise 2.2
• NCERT Solutions

Important Topics under NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Ex 2.2)

Chapter 2 of the class 9 maths syllabus is on Polynomials. It is a very important chapter that is covered in class 9 and is divided into 8 major topics. Students are advised to learn from ad practice the solved questions time and again to be able to master the concept of polynomials.

The following list has been provided to give the students a glimpse at the important topics that are covered under the chapter on Polynomials.

Classification of polynomials

Degree of a polynomial

Types of polynomial based on the degree

Constant polynomial

Linear polynomial

Quadratic polynomial

Cubic polynomial

Types of polynomial based on terms

Value of a polynomial

Zero of a polynomial

Operations on polynomials

Importance of Polynomials in Class 9

Polynomials are expressions having more than two algebraic terms. They can also be defined as the sum of several terms where the same variable/variables has/have different powers. 

Polynomials are practically a language used in most mathematical expressions and are used to represent appropriate relations between different variables or numbers. Students are encouraged to gather as much information from this chapter to be able to solve tricky problems based on polynomials easily in exams.

NCERT Solutions for Class 9 Maths Chapter 2 teach you about polynomials. Vedantu’s NCERT Solutions for Class 9 Maths Chapter 2 PDF makes you aware of every aspect of Polynomials - be it finding value, using algebraic identities or factorization of polynomials. Our Exercise 2.2 Class 9 Maths solution explains everything about polynomials in an easy to understand way. Vedantu provides students with a Free PDF download option for all the NCERT Book Solutions of updated CBSE textbooks . Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 9 Science , Maths solutions, and solutions of other subjects that are available on Vedantu only.

Access NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials

Exercise 2.2

1. Find the value of the polynomial $\text{5x - 4}{{\text{x}}^{\text{2}}}\text{ + 3}$  at 

i) $\text{x = 0}$

Ans: Let $\text{p}\left( \text{x} \right)\text{ = 5x - 4}{{\text{x}}^{\text{2}}}\text{ + 3}$

We will simply put the value of $\text{x}$ in the given polynomial.

$\Rightarrow \text{p}\left( \text{0} \right)\text{ = 5}\left( \text{0} \right)\text{ - 4}\left( {{\text{0}}^{\text{2}}} \right)\text{ + 3}$

$\Rightarrow \text{p}\left( \text{0} \right)\text{ = 3}$

Hence, the value of the polynomial at $\text{x = 0}$ is $\text{3}$.

ii) $\text{x = -1}$ 

$\Rightarrow \text{p}\left( \text{-1} \right)\text{ = 5}\left( \text{-1} \right)\text{ - 4}\left[ {{\left( \text{-1} \right)}^{\text{2}}} \right]\text{ + 3}$

$\Rightarrow \text{p}\left( \text{-1} \right)\text{ = - 5 - 4 + 3}$

$\Rightarrow \text{p}\left( \text{-1} \right)\text{ = - 6}$

Hence, the value of the polynomial at $\text{x = -1}$ is $\text{-6}$.

iii) $\text{x = 2}$ 

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 5}\left( \text{2} \right)\text{ - 4}\left( {{\text{2}}^{\text{2}}} \right)\text{ + 3}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 10 - 16 + 3}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = - 3}$

Hence, the value of the polynomial at $\text{x = 2}$ is $\text{-3}$.

2. Find $\text{p}\left( \text{0} \right)$, $\text{p}\left( \text{1} \right)$  and $\text{p}\left( \text{2} \right)$  for each of the following polynomials:

i) $\text{p}\left( \text{y} \right)\text{ = }{{\text{y}}^{\text{2}}}\text{ - y + 1}$ 

$\text{p}\left( \text{0} \right)\text{ = }{{\text{0}}^{\text{2}}}\text{ - 0 + 1}$.

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = 1}$.

$\text{p}\left( \text{1} \right)\text{ = }{{\text{1}}^{\text{2}}}\text{ - 1 + 1 = 1}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 1}$.

$\text{p}\left( \text{2} \right)\text{ = }{{\text{2}}^{\text{2}}}\text{ - 2 + 1}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 4 - 2 + 1}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 3}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 3}$.

ii) $\text{p}\left( \text{t} \right)\text{ = 2 + t + 2}{{\text{t}}^{\text{2}}}\text{ - }{{\text{t}}^{\text{3}}}$

$\text{p}\left( \text{0} \right)\text{ = 2 + 0 + 2}\left( {{\text{0}}^{\text{2}}} \right)\text{ - }\left( {{\text{0}}^{\text{3}}} \right)\text{ = 2}$

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = 2}$.

$\text{p}\left( \text{1} \right)\text{ = 2 + 1 + 2}\left( {{\text{1}}^{\text{2}}} \right)\text{ - }\left( {{\text{1}}^{\text{3}}} \right)\text{ = 4}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 4}$.

$\text{p}\left( \text{2} \right)\text{ = 2 + 2 + 2}\left( {{\text{2}}^{\text{2}}} \right)\text{ - }\left( {{\text{2}}^{\text{3}}} \right)$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 4 + 8 - 8}$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = 4}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 4}$.

iii) $\text{p}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{3}}}$

$\text{p}\left( \text{0} \right)\text{ = }{{\text{0}}^{\text{3}}}\text{ = 0}$

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = 0}$.

$\text{p}\left( \text{1} \right)\text{ = }{{\text{1}}^{\text{3}}}\text{ = 1}$

$\text{p}\left( \text{2} \right)\text{ = }{{\text{2}}^{\text{3}}}\text{ = 8}$

So, we get the value of $\text{p}\left( \text{2} \right)\text{ = 8}$.

iv) $\text{p}\left( \text{x} \right)\text{ =}\left( \text{x - 1} \right)\left( \text{x + 1} \right)$

$\text{p}\left( \text{0} \right)\text{ =}\left( \text{0 - 1} \right)\left( \text{0 + 1} \right)\text{ = -1}$

So, we get the value of $\text{p}\left( \text{0} \right)\text{ = -1}$.

$\text{p}\left( \text{1} \right)\text{ =}\left( \text{1 - 1} \right)\left( \text{1 + 1} \right)\text{ = 0}$

So, we get the value of $\text{p}\left( \text{1} \right)\text{ = 0}$.

$\text{p}\left( \text{2} \right)\text{ =}\left( \text{2 - 1} \right)\left( \text{2 + 1} \right)$

$\Rightarrow \text{p}\left( \text{2} \right)\text{ = }\left( \text{1} \right)\left( \text{3} \right)\text{ = 3}$

3. Verify whether the following are zeroes of the polynomial, indicated against them.

i) $\text{p}\left( \text{x} \right)\text{ = 3x + 1}$,$\text{x = -}\frac{\text{1}}{\text{3}}$

Ans: We are given: $\text{x = -}\frac{\text{1}}{\text{3}}$. If it is the zero of the polynomial $\text{p}\left( \text{x} \right)\text{ = 3x + 1}$, then $\text{p}\left( \text{-}\frac{\text{1}}{\text{3}} \right)$ should be  $\text{0}$.

$\text{p}\left(\text{-}\frac{\text{1}}{\text{3}}\right)\text{=3}\left(\text{-}\frac{\text{1}}{\text{3}} \right)\text{ + 1 = -1 + 1 = 0}$

Hence, we can say that $\text{x = -}\frac{\text{1}}{\text{3}}$ is a zero of the given polynomial.

ii) $\text{p}\left( \text{x} \right)\text{= 5x - }\!\!\pi\!\!\text{ }$ ,$\text{x = }\frac{\text{4}}{\text{5}}$

Ans: We are given: $\text{x = }\frac{\text{4}}{\text{5}}$. If it is the zero of the polynomial $\text{p}\left( \text{x} \right)\text{ = 5x -  }\text{ }\!\!\pi\!\!\text{ }$, then $\text{p}\left( \frac{\text{4}}{\text{5}} \right)$ should be  $\text{0}$.

$\text{p}\left( \frac{\text{4}}{\text{5}} \right)\text{ = 5}\left( \frac{\text{4}}{\text{5}} \right)\text{ - 3}\text{.14 = 4 - 3}\text{.14 }\ne \text{ 0}$

Hence, we can say that $\text{x = }\frac{\text{4}}{\text{5}}$ is not a zero of the given polynomial.

iii) $\text{p(x) = }{{\text{x}}^{\text{2}}}\text{-1, x = 1, -1}$

Ans: We are given: $\text{x = 1}$  and  $\text{x = -1}$ . 

If they are zeros of polynomial  $\text{p(x) = }{{\text{x}}^{\text{2}}}\text{-1}$ , then $\text{p}\left( \text{1} \right)$ and $\text{p}\left( \text{-1} \right)$ should both be $\text{0}$.

$\text{p}\left( \text{1} \right)\text{ = }{{\left( \text{1} \right)}^{\text{2}}}\text{-1 = 0}$

$\text{p}\left( \text{-1} \right)\text{ = }{{\left( \text{-1} \right)}^{\text{2}}}\text{-1 = 0}$

Hence, we can say that $\text{x = 1}$ and $\text{x = -1}$  are zeroes of the given polynomial.

iv) $\text{p(x) = (x+1)(x-2), x = -1, 2}$

Ans: We are given: $\text{x = -1}$ and $\text{x = 2}$. 

If they are zeroes of the polynomial $\text{p}\left( \text{x} \right)\text{ = }\left( \text{x+1} \right)\left( \text{x-2} \right)$ , then $\text{p}\left( \text{-1} \right)$ and $\text{p}\left( \text{2} \right)$ should be $\text{0}$.

$\text{p}\left( \text{-1} \right)\text{ = }\left( \text{-1+1} \right)\left( \text{-1-2} \right)\text{ = }\left( \text{0} \right)\left( \text{-3} \right)\text{ = 0}$

$\text{p}\left( \text{2} \right)\text{ = }\left( \text{2+1} \right)\left( \text{2-2} \right)\text{ = }\left( \text{3} \right)\left( \text{0} \right)\text{ = 0}$

Hence, we can say that $\text{x = -1}$ and $\text{x = 2}$  are zeroes of the given polynomial.

v) $\text{p(x) = }{{\text{x}}^{\text{2}}}\text{, x = 0}$

Ans: We are given $\text{x = 0}$. 

If it is a zero of the polynomial $\text{p}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{2}}}$  , then $\text{p}\left( \text{0} \right)$ should be $\text{0}$.

$\text{ p}\left( \text{0} \right)\text{ = }{{\left( \text{0} \right)}^{\text{2}}}\text{ = 0}$

Hence, we can say that $\text{x = 0}$ is a zero of the given polynomial.

vi) $\text{p(x) = lx + m, x = -}\frac{\text{m}}{\text{l}}$

Ans: We are given: $\text{x = -}\frac{\text{m}}{\text{l}}$. 

If it is a zero of the polynomial  $\text{p}\left( \text{x} \right)\text{ = lx + m}$ , then $\text{p}\left( \text{-}\frac{\text{m}}{\text{l}} \right)$ should be  $\text{0}$ .

Here, $\text{p}\left( \text{-}\frac{\text{m}}{\text{l}} \right)\text{ = l}\left( \text{-}\frac{\text{m}}{\text{l}} \right)\text{ + m = -m + m = 0}$

Hence, we can say that $\text{x = -}\frac{\text{m}}{\text{l}}$  is a zero of the given polynomial.

vii)$\text{p(x)=3}{{\text{x}}^{\text{2}}}\text{-1,x=-}\frac{\text{1}}{\sqrt{\text{3}}}\text{,}\frac{\text{2}}{\sqrt{\text{3}}}$

Ans: We are given:$\text{x = }\frac{\text{-1}}{\sqrt{\text{3}}}$  and  $\text{x = }\frac{\text{2}}{\sqrt{\text{3}}}$. 

If they are zeroes of the polynomial $\text{p}\left( \text{x} \right)\text{ = 3}{{\text{x}}^{\text{2}}}\text{-1}$, then $\text{p}\left( \frac{\text{-1}}{\sqrt{\text{3}}} \right)$and $\text{p}\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)$   should be $\text{0}$ .

$\text{p}\left( \frac{\text{-1}}{\sqrt{\text{3}}} \right)\text{ = 3}{{\left( \frac{\text{-1}}{\sqrt{\text{3}}} \right)}^{\text{2}}}\text{-1 = 3}\left( \frac{\text{1}}{\text{3}} \right)\text{-1 = 1-1 = 0}$

$\text{p}\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)\text{ = 3}{{\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)}^{\text{2}}}\text{-1 = 3}\left( \frac{\text{4}}{\text{3}} \right)\text{-1 = 4-1 = 3}$

Hence, we can say that $\text{x = }\frac{\text{-1}}{\sqrt{\text{3}}}$ is a zero of the given polynomial.

However, the value of $\text{x = }\frac{\text{2}}{\sqrt{\text{3}}}$ is not a zero of the given polynomial.

viii) $\text{p(x) = 2x+1, x = }\frac{\text{1}}{\text{2}}$

Ans: We are given: $\text{x = }\frac{\text{1}}{\text{2}}$. 

If it is a zero of polynomial $\text{p}\left( \text{x} \right)\text{ = 2x+1}$ , then $\text{p}\left( \frac{\text{1}}{\text{2}} \right)$ should be $\text{0}$

Here, $\text{p}\left( \frac{\text{1}}{\text{2}} \right)\text{ = 2}\left( \frac{\text{1}}{\text{2}} \right)\text{+1 = 1+1 = 2}$.

So, we get the value $\text{p}\left( \frac{\text{1}}{\text{2}} \right)\ne \text{0}$  .

Hence, we can say that $\text{x = }\frac{\text{1}}{\text{2}}$ is not a zero of the given polynomial.

4. Find the zero of the polynomial in each of the following cases:

i) $\text{p}\left( \text{x} \right)\text{ = x+5}$

Ans:  If $\text{x}$ is zero of the polynomial, then we can say that $\text{p}\left( \text{x} \right)\text{ = 0}$.

Let $\text{p}\left( \text{x} \right)\text{ = 0}$

$\Rightarrow \text{x+5 = 0}$

$\Rightarrow \text{x = -5}$

Therefore, $\text{x = -5}$ is a zero of the given polynomial.

ii) $\text{p}\left( \text{x} \right)\text{ = x-5}$ 

$\Rightarrow \text{x-5 = 0}$

$\Rightarrow \text{x = 5}$

Therefore, $\text{x = 5}$ is a zero of the given polynomial.

iii) $\text{p}\left( \text{x} \right)\text{ = 2x+5}$

$\Rightarrow \text{2x+5 = 0}$

$\Rightarrow \text{x = -}\frac{\text{5}}{\text{2}}$

Therefore, $\text{x = -}\frac{\text{5}}{\text{2}}$ is a zero of the given polynomial.

iv) $\text{p}\left( \text{x} \right)\text{ = 3x-2}$

$\Rightarrow \text{3x-2 = 0}$

$\Rightarrow \text{x = }\frac{\text{2}}{\text{3}}$

Therefore, $\text{x = }\frac{\text{2}}{\text{3}}$  is a zero of the given polynomial.

v) $\text{p}\left( \text{x} \right)\text{ = 3x}$

$\Rightarrow \text{3x = 0}$

$\Rightarrow \text{x = 0}$

Therefore, $\text{x = 0}$ is a zero of the given polynomial.

vi) $\text{p}\left( \text{x} \right)\text{ = ax, a}\ne \text{0}$

$\Rightarrow \text{ax = 0}$

It is also given that $\text{a}$ is non-zero. 

vii) $\text{p}\left( \text{x} \right)\text{ = cx+d, c}\ne \text{0, c, d}$ are real numbers.

$\Rightarrow \text{cx+d = 0}$

It is also given that $\text{c}$ is non-zero.

$\Rightarrow \text{x = -}\frac{\text{d}}{\text{c}}$

Therefore, $\text{x = -}\frac{\text{d}}{\text{c}}$  is a zero of the given polynomial.

Class 9 Maths Chapter 2 - Exercise 2.2

Vedantu’s Class 9 Maths Chapter 2 Exercise 2.2 includes the following problems from polynomial-

Finding the value of polynomial when the value of the variable is randomly fixed by the teacher,For example, if I ask you to find the value of 2x + 3x when I have randomly set the value of X as 5, you will put the value - 5 in place of X and then multiply 2 and 3 with 5.

Finding whether a particular value of the variable will result in the polynomial to be equated as  For example, if X equals to 5 in the equation 2X - 10, then the X = 5 is a zero of the given polynomial.

Finding which value of the variable will result in the polynomial to be equated as 0. For example, if I want to make the value of 3X+1-7 then I have to assign the value of X as 2.

What are Polynomials? Constants and Variables

Before discussing polynomials, it is necessary that you have a clear idea about variables and exponents. Consider the following equation -

2x + 3x = 10

Here we don't know at first what X is. When we solve the equation, we come to the conclusion that here X is 2. So, what X does here is it answers the ‘what’ of a question. In this equation, the question that is implied is - what should we multiply 2 and 3 with so that the addition of the product results in 10?

Again when we are confronted with the following equation - 

2x + 3x = 20

the result of X changes. So, the value of X varies the situation.

This is what we call variable. This variable is generally denoted by x,y,z etc.

Again, if you look at the above two examples, the values of 2 and 3 never change. In both situations, the values of 2 and 3 remain fixed. These are called constants.

Now, look at this,

Here the value of 2 increases exponentially ( 2*2*2). The power 3 is known as an exponent.

If you look at the algebraic expression - 2x + 3x - written above, you will see there are two parts in it - 2x and 3x. These parts are called terms.

Polynomial is a mathematical expression that has multiple terms and consists of constants, variables and non-negative or non-fractionated exponents. The polynomial will only involve addition, subtraction or multiplication.

So 2x + 3x is a polynomial in the variable X.

Again, 2³ + X³ + 3² is a polynomial.

But X + 1∕X is not a polynomial since 1/x can be written as . As said earlier, a polynomial does not allow negative exponents.

Expressions that have √X are not polynomials too since √X can be written as and we know that this is not accepted by polynomials.

If you look at the expression 2x + 3x, the numbers before the variables are called coefficients. 

Class 9 Maths Chapter 2 – Other Exercises (2.1 and 2.3)

Here are a few other exercises that you will find in the NCERT Solutions for Class 9 Maths Chapter 2 PDF – 

Exercise 2.1  

In this exercise of Class 9 Maths Chapter 2, students will be introduced to the concept of polynomials. They will obtain an indent knowledge regarding polynomials in one variable and how to solve the associated sums through shortcut techniques. Here are some important questions for exams that are provided in the exercise.

Question 1 : Finding the polynomials among the expressions in one variable and stating reasons for the answers.

Question 2 : Finding the coefficients of x2 in the provided expressions.

Question 3 : Examples of a binomial of degree 35 and monomial of degree 100.

Question 4: Determining the degree of polynomials.

Question 5 : Classifying linear, quadratic and cubic polynomial expressions.

Exercise 2.3

This chapter is a continuation of Exercise 2.2 Class 9 of NCERT maths book. Students will get in-depth knowledge about the Factor Theorem by solving all questions provided in this exercise step by step. 

Question 1 : Finding the remainder when divided by expressions provided.

Question 2: Determining remainders. 

Question 3: Checking if provided expressions are factors.

How Can Vedantu Help You?

Vedantu's Polynomials Class 9 solutions will help you understand all the topics included in the Polynomials chapter. NCERT guidelines specifically advise teachers to build the fundamental knowledge of the subjects in the students. Vedantu follows this NCERT advice thoroughly. That is why you will find not only solutions to the questions on polynomials but also the explanations on the logic behind our solutions. For example - in question number 2 of Ex 2.1 of our Polynomials Class 9 NCERT Solutions when we are asked to find the coefficients of given polynomials we showed how has an invisible 1 before it. So the coefficient of X² is 1.

Our answers in Class 9th Maths NCERT Solutions Chapter 2 include every step. No step has been skipped. You will not struggle to understand how we reached the result from a given equation. So in question 1(i) of Ex 2.2, we showed how we arrived at the answer 3 by solving the equation step by step. This will not only help the students to understand the granularities in the chapter but will also help them to get good marks.

Every Polynomial formula Class 9 has been used in our solution. The students will get to learn all the aspects of the chapter. Reading our PDF, a student can answer any question from this chapter with confidence.

The Polynomials Class 9 PDF is written in a lucid language. Apart from the Maths expressions, we have used simple, easy to understand words so that students can understand the solutions. Even the math expressions are adequately explained wherever deemed fit.

The Class 9 Maths Chapter 2 PDF has accurate solutions to every polynomials problem. The answers are written by expert teachers who know how to solve a Maths problem so that students understand the process and even the examiners get impressed by the detailed process. These teachers are well aware of NCERT guidelines and follow them to write their answers. There is no unnecessary information in our solutions.

Vedantu’s Polynomial Class 9 PDF is neatly organised. The solutions are written in an uncluttered, easy to understand way. That is why in question 1 of Ex2.3, you will find that we have used red ink to indicate the cutting off of -2x² and +2x². This visually pleasurable way of reading the solutions will help the students to maintain their focus on the solutions.

You can use our NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 to learn all the intricacies of polynomials. These solutions can provide you with everything that you need to excel in the chapter. The answers in our PDF are not written by part-time bloggers, but they are written by professional teachers who took care to make the solutions as helpful as possible.

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NCERT Solutions for Class 9 Maths

Chapter 1 - Number System

Chapter 3 - Coordinate Geometry

Chapter 4 - Linear Equations in Two Variables

Chapter 5 - Introductions to Euclid's Geometry

Chapter 6 - Lines and Angles

Chapter 7 - Triangles

Chapter 8 - Quadrilaterals

Chapter 9 - Areas of Parallelogram and Triangles

Chapter 10 - Circles

Chapter 11 - Constructions

Chapter 12 - Heron's formula

Chapter 13 - Surface area and Volumes

Chapter 14 - Statistics

Chapter 15 - Probability

FAQs on NCERT Solutions for Class 9 Maths Chapter 2: Polynomials - Exercise 2.2

1.  Give a brief description of the topic.

Polynomial is acquired from the word “poly” which means “many” and the word “nominal” refers to “term”. In Math subject, a polynomial expression consists of variables which are also known as coefficients and indeterminates. The coefficients require the operations of addition, subtraction, non-negative integer exponents of variables and multiplication. 

Polynomials are utterances with one or more phrases with a non-zero coefficient. A polynomial can also have one or more than one number of terms. In the form of a polynomial, each utterance in it is known as a term.  Suppose x 2 + 4x+ 2 is polynomial, then the expressions x 2 , 4x, and 2 are the terms of the polynomial.  Each term of the polynomial has a coefficient. For example, if 4x+1 is the polynomial, then the coefficient of x is 4.

2. What are the important terms of a polynomial?

A term might either be a variable or a single digit or it also could be a sequence of the variable with digits.

The degree of the polynomial is the most distinguished power of the variable in a polynomial.

A polynomial of degree 1 will be termed as a linear polynomial.

A polynomial of degree 2 will be described as a quadratic polynomial.

A polynomial of degree 3 is termed as a cubic polynomial.

A polynomial of 1 term is termed as a monomial.

A polynomial of 2 terms is termed as binomial.

A polynomial of 3 terms is termed as a trinomial.

A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0, where a is also termed as root of the equation p(x) = 0.

A linear polynomial in one variable holds an individual zero, a polynomial of a non-zero constant has no zero, and every real character is a zero of the zero polynomial.

3. What are the topics that are covered in this chapter?

Polynomial is an algebraic expression which includes constants, variables and exponents. It is the expression where the variables have just certain elemental powers. The topics that are covered in this chapter are:

Introduction

Polynomials in One Variable

Zeros of Polynomials

Remainder Theorem

Factorisation of Polynomials

Algebraic Identities

Polynomial expressions are algebraic equations which will have two or more terms with the equal variables of different exponents. This is one of the notable topics for class 9 Mathematics, which students require to learn in order to gain profound knowledge about complex algebraic expressions. Mathematics is one of those subjects that serve as a foundation of numbers, analysis, figures and logic.

4. Are the answers provided by Vedantu, sufficient to attain accurate marks?

Our answers are made to the point and they are drafted to aid you from the exam portion of the view. Answers to the exercising questions are certainly given with examples and they are 100% curate. Our answers will make your learning easy for the exam as they are fitted to be compatible with the tips given with the assistance of using CBSE maths Syllabus and NCERT Book.

Our answers will assist you in advancing a conceptual basis with all the principal ideas in a comprehensible language. The exercising covers all of the vital topics and subtopics of the chapter which might occur for your Class 9 maths exams. You also can clear all of your doubts from here and in this manner, you can definitely perform well in your Class 9 maths exam.

5. Do I need to practice all the questions given in the NCERT Solutions for Class 9 Maths Chapter 2?

The NCERT textbook is the best textbook to prepare for the final exams as it is where most of the questions are asked in the final exam. The exercises and the examples are explained in a simpler manner for all students to understand. Important concepts are also explained in the PDFs. It is important that you practice all the questions and concepts while preparing for the exam. You should not leave any questions. Being thorough with all the questions will help you score good marks in the final exam. You can visit the Vedantu website or download the Vedantu app to access these resources at free of cost.

6. How many exercises are there in this chapter?

Chapter 2 of Maths from the NCERT textbook has three exercises- 2.1, 2.2 and 2.3. Each exercise and its solution is explained step by step in the NCERT Solutions for Class 9 Maths Chapter 2. Not a single question or step has been skipped where they are explained. The examples from within the chapter are also explained. The solutions have been written by professionals with the aim of making it easier for all the students depending on their calibre to understand. 

7. How many questions are there in each exercise?

There are a total of three exercises in the Chapter 2 of Class 10 NCERT Maths textbook. Exercise 2.1 has five questions that cover introduction to polynomials and linear and quadratic linear expressions. Exercise 2.2 has four questions that mainly concentrate on finding the value of the polynomials. The last exercise is 2.3 which focuses on the factor theorem. The questions ask to find the remainders and their division. 

8. What are constants and variables?

Constants are the numeric values attached to a variable (x,y) in any equation. They stay the same and do not change. They help you determine the value of the variable. Variables are the X or Y attached to a constant whose value needs to be determined. It may vary in different situations.  It is a result of the constants and the answer on the other side that helps determine the value of the variables.

9. Is Class 9 Maths NCERT Solutions enough to revise for the final exam?

The NCERT Solutions for each subject and chapter provides summaries and revision notes for the students to refer to and prepare from. These are sufficient as they cover every aspect of the chapter and do not exclude any topic. The students can take the help of the Class 9 Maths NCERT Solutions to prepare for the exam and revise by going through the revision notes provided. The formulas and concepts are explained in an easy and systematic manner. 

NCERT Solutions for Class 9

Cbse study materials for class 9, cbse study materials.

• NCERT Solutions
• NCERT Class 9
• NCERT 9 Maths
• Chapter 2: Polynomial
• Exercise 2.5

NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials Exercise 2.5

NCERT Solutions Class 9 Maths Chapter 2 – Polynomials Exercise 2.5 are given here. These NCERT Maths solutions are created by our subject experts, who make it easy for students to learn. The students use NCERT Class 9 Maths Solutions for reference while solving the exercise problems. The fifth exercise in Polynomials- Exercise 2.5 discusses the Algebraic Identities.

The experts provide a detailed and stepwise explanation of each answer to the questions given in the exercises in the NCERT textbook for Class 9. The NCERT solutions are always prepared by following NCERT guidelines so that it should cover the whole syllabus accordingly. These are very helpful in scoring well in board examinations.

NCERT Solutions for Class 9 Maths Chapter 2- Polynomials Exercise 2.5

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Access Other Exercise Solutions of Class 9 Maths Chapter 2- Polynomials

The Exercise wise links for NCERT Class 9 Solutions Maths Chapter 2 are given below.

Exercise 2.1 Solutions 5 Questions Exercise 2.2 Solutions 4 Questions Exercise 2.3 Solutions 3 Questions Exercise 2.4 Solutions 5 Questions

Access Answers to NCERT Class 9 Maths Chapter 2 – Polynomials Exercise 2.5

1. Use suitable identities to find the following products:

(i) (x+4)(x +10) 

Using the identity, (x+a)(x+b) = x 2 +(a+b)x+ab

(x+4)(x+10) = x 2 +(4+10)x+(4×10)

= x 2 +14x+40

(ii) (x+8)(x –10)     

(x+8)(x−10) = x 2 +(8+(−10))x+(8×(−10))

= x 2 +(8−10)x–80

= x 2 −2x−80

(iii) (3x+4)(3x–5)

(3x+4)(3x−5) = (3x) 2 +[4+(−5)]3x+4×(−5)

= 9x 2 +3x(4–5)–20

= 9x 2 –3x–20

(iv) (y 2 +3/2)(y 2 -3/2)

Using the identity, (x+y)(x–y) = x 2 –y 2

(y 2 +3/2)(y 2 –3/2) = (y 2 ) 2 –(3/2) 2

2. Evaluate the following products without multiplying directly:

(i) 103×107

103×107= (100+3)×(100+7)

Using identity, [(x+a)(x+b) = x 2 +(a+b)x+ab

Here, x = 100

We get, 103×107 = (100+3)×(100+7)

= (100) 2 +(3+7)100+(3×7)

= 10000+1000+21

(ii) 95×96  

95×96 = (100-5)×(100-4)

Using identity, [(x-a)(x-b) = x 2 -(a+b)x+ab

We get, 95×96 = (100-5)×(100-4)

= (100) 2 +100(-5+(-4))+(-5×-4)

= 10000-900+20

(iii) 104×96

104×96 = (100+4)×(100–4)

Using identity, [(a+b)(a-b)= a 2 -b 2 ]

Here, a = 100

We get, 104×96 = (100+4)×(100–4)

= (100) 2 –(4) 2

3. Factorize the following using appropriate identities:

(i) 9x 2 +6xy+y 2

9x 2 +6xy+y 2 = (3x) 2 +(2×3x×y)+y 2

Using identity, x 2 +2xy+y 2 = (x+y) 2

Here, x = 3x

= (3x+y)(3x+y)

(ii) 4y 2 −4y+1

4y 2 −4y+1 = (2y) 2 –(2×2y×1)+1

Using identity, x 2 – 2xy + y 2 = (x – y) 2

Here, x = 2y

4y 2 −4y+1 = (2y) 2 –(2×2y×1)+1 2

= (2y–1)(2y–1)

(iii)  x 2 –y 2 /100

x 2 –y 2 /100 = x 2 –(y/10) 2

Using identity, x 2 -y 2 = (x-y)(x+y)

Here, x = x

= (x–y/10)(x+y/10)

4. Expand each of the following, using 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

(vi) ((1/4)a-(1/2)b +1) 2

Using identity, (x+y+z) 2 = x 2 +y 2 +z 2 +2xy+2yz+2zx

(x+2y+4z) 2 = x 2 +(2y) 2 +(4z) 2 +(2×x×2y)+(2×2y×4z)+(2×4z×x)

= x 2 +4y 2 +16z 2 +4xy+16yz+8xz

(ii) (2x−y+z) 2  

Here, x = 2x

(2x−y+z) 2 = (2x) 2 +(−y) 2 +z 2 +(2×2x×−y)+(2×−y×z)+(2×z×2x)

= 4x 2 +y 2 +z 2 –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)

= 4x 2 +9y 2 +4z 2 –12xy+12yz–8xz

Using identity (x+y+z) 2 = x 2 +y 2 +z 2 +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)

= 9a 2 + 49b 2 + c 2 – 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)

= 4x 2 +25y 2 +9z 2 – 20xy–30yz+12zx

(vi) ((1/4)a-(1/2)b+1) 2

Here, x = (1/4)a

y = (-1/2)b

5. Factorize:

(i) 4x 2 +9y 2 +16z 2 +12xy–24yz–16xz

(ii ) 2x 2 +y 2 +8z 2 –2√2xy+4√2yz–8xz

We can say that, x 2 +y 2 +z 2 +2xy+2yz+2zx = (x+y+z) 2

4x 2 +9y 2 +16z 2 +12xy–24yz–16xz = (2x) 2 +(3y) 2 +(−4z) 2 +(2×2x×3y)+(2×3y×−4z)+(2×−4z×2x)

= (2x+3y–4z) 2

= (2x+3y–4z)(2x+3y–4z)

(ii) 2x 2 +y 2 +8z 2 –2√2xy+4√2yz–8xz

Using identity, (x +y+z) 2 = x 2 +y 2 +z 2 +2xy+2yz+2zx

2x 2 +y 2 +8z 2 –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)

6. Write the following cubes in expanded form:

(i) (2x+1) 3

(ii) (2a−3b) 3

(iii) ((3/2)x+1) 3

(iv) (x−(2/3)y) 3

Using identity,(x+y) 3 = x 3 +y 3 +3xy(x+y)

(2x+1) 3 = (2x) 3 +1 3 +(3×2x×1)(2x+1)

= 8x 3 +1+6x(2x+1)

= 8x 3 +12x 2 +6x+1

Using identity,(x–y) 3 = x 3 –y 3 –3xy(x–y)

(2a−3b) 3 = (2a) 3 −(3b) 3 –(3×2a×3b)(2a–3b)

= 8a 3 –27b 3 –18ab(2a–3b)

= 8a 3 –27b 3 –36a 2 b+54ab 2

((3/2)x+1) 3 =((3/2)x) 3 +1 3 +(3×(3/2)x×1)((3/2)x +1)

(iv)  (x−(2/3)y) 3

Using identity, (x –y) 3 = x 3 –y 3 –3xy(x–y)

7. Evaluate the following using suitable identities: 

(ii) (102) 3

(iii) (998) 3

We can write 99 as 100–1

(99) 3 = (100–1) 3

= (100) 3 –1 3 –(3×100×1)(100–1)

= 1000000 –1–300(100 – 1)

= 1000000–1–30000+300

We can write 102 as 100+2

(100+2) 3 =(100) 3 +2 3 +(3×100×2)(100+2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

We can write 99 as 1000–2

(998) 3 =(1000–2) 3

=(1000) 3 –2 3 –(3×1000×2)(1000–2)

= 1000000000–8–6000(1000– 2)

= 1000000000–8- 6000000+12000

= 994011992

8. Factorise each of the following:

(i) 8a 3 +b 3 +12a 2 b+6ab 2

(ii) 8a 3 –b 3 –12a 2 b+6ab 2

(iii) 27–125a 3 –135a +225a 2    

(iv) 64a 3 –27b 3 –144a 2 b+108ab 2

(v) 27p 3 –(1/216)−(9/2) p 2 +(1/4)p

The expression, 8a 3 +b 3 +12a 2 b+6ab 2 can be written as (2a) 3 +b 3 +3(2a) 2 b+3(2a)(b) 2

8a 3 +b 3 +12a 2 b+6ab 2 = (2a) 3 +b 3 +3(2a) 2 b+3(2a)(b) 2

= (2a+b)(2a+b)(2a+b)

Here, the identity, (x +y) 3 = x 3 +y 3 +3xy(x+y) is used.

The expression, 8a 3 –b 3 −12a 2 b+6ab 2 can be written as (2a) 3 –b 3 –3(2a) 2 b+3(2a)(b) 2

8a 3 –b 3 −12a 2 b+6ab 2 = (2a) 3 –b 3 –3(2a) 2 b+3(2a)(b) 2

= (2a–b)(2a–b)(2a–b)

Here, the identity,(x–y) 3 = x 3 –y 3 –3xy(x–y) is used.

(iii) 27–125a 3 –135a+225a 2  

The expression, 27–125a 3 –135a +225a 2 can be written as 3 3 –(5a) 3 –3(3) 2 (5a)+3(3)(5a) 2

27–125a 3 –135a+225a 2 = 3 3 –(5a) 3 –3(3) 2 (5a)+3(3)(5a) 2

= (3–5a)(3–5a)(3–5a)

Here, the identity, (x–y) 3 = x 3 –y 3 -3xy(x–y) is used.

The expression, 64a 3 –27b 3 –144a 2 b+108ab 2 can be written as (4a) 3 –(3b) 3 –3(4a) 2 (3b)+3(4a)(3b) 2

64a 3 –27b 3 –144a 2 b+108ab 2 = (4a) 3 –(3b) 3 –3(4a) 2 (3b)+3(4a)(3b) 2

=(4a–3b)(4a–3b)(4a–3b)

Here, the identity, (x – y) 3 = x 3 – y 3 – 3xy(x – y) is used.

(v) 27p 3 – (1/216)−(9/2) p 2 +(1/4)p

The expression, 27p 3 –(1/216)−(9/2) p 2 +(1/4)p can be written as

(3p) 3 –(1/6) 3 −(9/2) p 2 +(1/4)p = (3p) 3 –(1/6) 3 −3(3p)(1/6)(3p – 1/6)

Using (x – y) 3 = x 3 – y 3 – 3xy (x – y)

27p 3 –(1/216)−(9/2) p 2 +(1/4)p = (3p) 3 –(1/6) 3 −3(3p)(1/6)(3p – 1/6)

Taking x = 3p and y = 1/6

= (3p–1/6) 3

= (3p–1/6)(3p–1/6)(3p–1/6)

(i) x 3 +y 3 = (x+y)(x 2 –xy+y 2 )

(ii) x 3 –y 3 = (x–y)(x 2 +xy+y 2 )

We know that, (x+y) 3 = x 3 +y 3 +3xy(x+y)

⇒ x 3 +y 3 = (x+y) 3 –3xy(x+y)

⇒ x 3 +y 3 = (x+y)[(x+y) 2 –3xy]

Taking (x+y) common ⇒ x 3 +y 3 = (x+y)[(x 2 +y 2 +2xy)–3xy]

⇒ x 3 +y 3 = (x+y)(x 2 +y 2 –xy)

(ii) x 3 –y 3 = (x–y)(x 2 +xy+y 2 ) 

We know that,(x–y) 3 = x 3 –y 3 –3xy(x–y)

⇒ x 3 −y 3 = (x–y) 3 +3xy(x–y)

⇒ x 3 −y 3 = (x–y)[(x–y) 2 +3xy]

Taking (x+y) common ⇒ x 3 −y 3 = (x–y)[(x 2 +y 2 –2xy)+3xy]

⇒ x 3 +y 3 = (x–y)(x 2 +y 2 +xy)

10. Factorize each of the following:

(i) 27y 3 +125z 3

(ii) 64m 3 –343n 3

The expression, 27y 3 +125z 3 can be written as (3y) 3 +(5z) 3

27y 3 +125z 3 = (3y) 3 +(5z) 3

We know that, x 3 +y 3 = (x+y)(x 2 –xy+y 2 )

= (3y+5z)[(3y) 2 –(3y)(5z)+(5z) 2 ]

= (3y+5z)(9y 2 –15yz+25z 2 )

The expression, 64m 3 –343n 3 can be written as (4m) 3 –(7n) 3

64m 3 –343n 3 = (4m) 3 –(7n) 3

We know that, x 3 –y 3 = (x–y)(x 2 +xy+y 2 )

= (4m–7n)[(4m) 2 +(4m)(7n)+(7n) 2 ]

= (4m–7n)(16m 2 +28mn+49n 2 )

11. Factorise: 27x 3 +y 3 +z 3 –9xyz 

The expression 27x 3 +y 3 +z 3 –9xyz can be written as (3x) 3 +y 3 +z 3 –3(3x)(y)(z)

27x 3 +y 3 +z 3 –9xyz  = (3x) 3 +y 3 +z 3 –3(3x)(y)(z)

We know that, x 3 +y 3 +z 3 –3xyz = (x+y+z)(x 2 +y 2 +z 2 –xy –yz–zx)

= (3x+y+z)[(3x) 2 +y 2 +z 2 –3xy–yz–3xz]

= (3x+y+z)(9x 2 +y 2 +z 2 –3xy–yz–3xz)

12. Verify that:

x 3 +y 3 +z 3 –3xyz = (1/2) (x+y+z)[(x–y) 2 +(y–z) 2 +(z–x) 2 ]

We know that,

x 3 +y 3 +z 3 −3xyz = (x+y+z)(x 2 +y 2 +z 2 –xy–yz–xz)

⇒ x 3 +y 3 +z 3 –3xyz = (1/2)(x+y+z)[2(x 2 +y 2 +z 2 –xy–yz–xz)]

= (1/2)(x+y+z)(2x 2 +2y 2 +2z 2 –2xy–2yz–2xz)

= (1/2)(x+y+z)[(x 2 +y 2 −2xy)+(y 2 +z 2 –2yz)+(x 2 +z 2 –2xz)]

= (1/2)(x+y+z)[(x–y) 2 +(y–z) 2 +(z–x) 2 ]

13. If  x+y+z = 0, show that x 3 +y 3 +z 3 = 3xyz.

x 3 +y 3 +z 3 -3xyz = (x +y+z)(x 2 +y 2 +z 2 –xy–yz–xz)

Now, according to the question, let (x+y+z) = 0,

then, x 3 +y 3 +z 3 -3xyz = (0)(x 2 +y 2 +z 2 –xy–yz–xz)

⇒ x 3 +y 3 +z 3 –3xyz = 0

⇒ x 3 +y 3 +z 3 = 3xyz

Hence Proved

14. Without actually calculating the cubes, find the value of each of the following:

(i) (−12) 3 +(7) 3 +(5) 3

(ii) (28) 3 +(−15) 3 +(−13) 3

Let a = −12

We know that if x+y+z = 0, then x 3 +y 3 +z 3 =3xyz.

Here, −12+7+5=0

(−12) 3 +(7) 3 +(5) 3 = 3xyz

= 3×-12×7×5

(28) 3 +(−15) 3 +(−13) 3

We know that if x+y+z = 0, then x 3 +y 3 +z 3 = 3xyz.

Here, x+y+z = 28–15–13 = 0

(28) 3 +(−15) 3 +(−13) 3 = 3xyz

= 0+3(28)(−15)(−13)

15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: 

(i) Area : 25a 2 –35a+12

(ii) Area : 35y 2 +13y–12

Using the splitting the middle term method,

We have to find a number whose sum = -35 and product =25×12=300

We get -15 and -20 as the numbers [-15+-20=-35 and -15×-20=300]

25a 2 –35a+12 = 25a 2 –15a−20a+12

= 5a(5a–3)–4(5a–3)

= (5a–4)(5a–3)

Possible expression for length  = 5a–4

Possible expression for breadth  = 5a –3

We have to find a number whose sum = 13 and product = 35×-12 = 420

We get -15 and 28 as the numbers [-15+28 = 13 and -15×28=420]

35y 2 +13y–12 = 35y 2 –15y+28y–12

= 5y(7y–3)+4(7y–3)

= (5y+4)(7y–3)

Possible expression for length  = (5y+4)

Possible expression for breadth  = (7y–3)

16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? 

(i) Volume : 3x 2 –12x

(ii) Volume : 12ky 2 +8ky–20k

3x 2 –12x can be written as 3x(x–4) by taking 3x out of both the terms.

Possible expression for length = 3

Possible expression for breadth = x

Possible expression for height = (x–4)

(ii) Volume: 12ky 2 +8ky–20k

12ky 2 +8ky–20k can be written as 4k(3y 2 +2y–5) by taking 4k out of both the terms.

12ky 2 +8ky–20k = 4k(3y 2 +2y–5)

= 4k(3y 2 +5y–3y–5)

= 4k[y(3y+5)–1(3y+5)]

= 4k(3y+5)(y–1)

Possible expression for length = 4k

Possible expression for breadth = (3y +5)

Possible expression for height = (y -1)

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