CBSE Class 9 Maths Chapter 2 Polynomials NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 9 PDF

This comprehensive set of NCERT Solutions for CBSE Class 9 Mathematics, Chapter 2: Polynomials, provides detailed explanations and step-by-step answers for all exercises. The chapter introduces the fundamental concepts of polynomials, including their definition, identification of polynomials in one variable, and distinguishing them from non-polynomial expressions. It covers how to find coefficients of terms, determine the degree of a polynomial, and classify polynomials as linear, quadratic, or cubic based on their degree. Furthermore, the solutions guide students on evaluating polynomials for given values of the variable and verifying zeroes of polynomials. These solutions are designed to reinforce understanding and aid students in mastering the concepts for effective exam preparation and revision.

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Quick info

Board	CBSE
Class	Class 9
Subject	Mathematics
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 2: Polynomials

Chapter summary

Chapter 2 of NCERT Class 9 Maths focuses on Polynomials. This solution set covers the definition of polynomials, identifying polynomials in one variable, and understanding why certain expressions are not polynomials. It details how to find coefficients, determine the degree of polynomials (linear, quadratic, cubic), and provides examples of monomials and binomials. The exercises also involve evaluating polynomials at specific points and verifying if given values are zeroes of the polynomial.

Learning outcomes

Understand the definition of a polynomial and identify polynomials in one variable.
Distinguish between polynomials and non-polynomial expressions based on variable exponents.
Identify and write the coefficients of terms in a polynomial.
Determine the degree of a polynomial.
Classify polynomials as linear, quadratic, or cubic.
Evaluate a polynomial for a given value of the variable.
Verify if a given value is a zero of a polynomial.

Topics covered

Paper topics

Definition of Polynomials
Polynomials in One Variable
Coefficients of Polynomials
Degree of Polynomials
Types of Polynomials (Linear, Quadratic, Cubic)
Monomials, Binomials, Trinomials
Evaluating Polynomials
Zeroes of a Polynomial

Important topics

Definition and identification of polynomials
Determining the degree of a polynomial
Classifying polynomials (linear, quadratic, cubic)
Evaluating polynomials at a given value
Verifying zeroes of polynomials

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Questions and Solutions

Question 1

Which of the following expressions are polynomials in one variable and which are not? State the reasons for your answer.

(i) $4x^2 - 3x + 7$

(ii) $y^2 + \sqrt{2}$

(iii) $3\sqrt{t} + t\sqrt{2}$

(iv) $y + \frac{2}{y}$

(v) $x^{10} + y^3 + t^{50}$

Solution:

To determine if an expression is a polynomial in one variable, we check if the variable has non-negative integer exponents.

(i) $4x^2 - 3x + 7$ : This is a polynomial in one variable, $x$ , because all exponents of $x$ (which are 2, 1, and 0 for the constant term) are non-negative integers.

(ii) $y^2 + \sqrt{2}$ : This is a polynomial in one variable, $y$ . The exponent of $y$ is 2, which is a non-negative integer. The term $\sqrt{2}$ is a constant coefficient.

(iii) $3\sqrt{t} + t\sqrt{2}$ : This is not a polynomial. The term $3\sqrt{t}$ can be written as $3t^{1/2}$ . Since the exponent $1/2$ is not a whole number, it is not a polynomial.

(iv) $y + \frac{2}{y}$ : This is not a polynomial. The term $\frac{2}{y}$ can be written as $2y^{-1}$ . Since the exponent $-1$ is not a whole number, it is not a polynomial.

(v) $x^{10} + y^3 + t^{50}$ : This is not a polynomial in *one* variable. It is a polynomial in three variables ( $x$ , $y$ , and $t$ ) because it contains more than one distinct variable.

Question 2

Write the coefficients of $x^2$ in each of the following:

(i) $2 + x^2 + x$

(ii) $2 - x^2 + x^3$

(iii) $\frac{\pi}{2}x^2 + x$

(iv) $\sqrt{2}x - 1$

Solution:

The coefficient of $x^2$ is the numerical factor multiplying the $x^2$ term.

(i) In the polynomial $2 + x^2 + x$ , the term with $x^2$ is $1 \cdot x^2$ . Therefore, the coefficient of $x^2$ is 1.

(ii) In the polynomial $2 - x^2 + x^3$ , the term with $x^2$ is $-1 \cdot x^2$ . Therefore, the coefficient of $x^2$ is -1.

(iii) In the polynomial $\frac{\pi}{2}x^2 + x$ , the term with $x^2$ is $\frac{\pi}{2} \cdot x^2$ . Therefore, the coefficient of $x^2$ is $\frac{\pi}{2}$ .

(iv) In the polynomial $\sqrt{2}x - 1$ , there is no term containing $x^2$ . This can be thought of as $0 \cdot x^2$ . Therefore, the coefficient of $x^2$ is 0.

Question 3

Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Solution:

A binomial is a polynomial with two terms, and a monomial is a polynomial with one term. The degree of a polynomial is the highest power of the variable.

Example of a binomial of degree 35: A binomial has two terms, and the highest power of the variable must be 35. An example is $x^{35} + 5$ .

Example of a monomial of degree 100: A monomial has one term, and the highest power of the variable must be 100. An example is $2y^{100}$ .

Question 4

Write the degree of each of the following polynomials:

(i) $5x^3 + 4x^2 + 7x$

(ii) $4 - y^2$

(iii) $5t - \sqrt{7}$

(iv) 3

Solution:

The degree of a polynomial is the highest power of the variable in the polynomial.

(i) In $5x^3 + 4x^2 + 7x$ , the powers of $x$ are 3, 2, and 1. The highest power is 3. So, the degree is 3.

(ii) In $4 - y^2$ , the powers of $y$ are 0 (for the constant term 4, which is $4y^0$ ) and 2. The highest power is 2. So, the degree is 2.

(iii) In $5t - \sqrt{7}$ , the powers of $t$ are 1 and 0 (for $\sqrt{7}$ , which is $\sqrt{7}t^0$ ). The highest power is 1. So, the degree is 1.

(iv) The polynomial 3 is a constant polynomial. It can be written as $3x^0$ . The highest power of $x$ is 0. So, the degree is 0.

Question 5

Classify the following as linear, quadratic and cubic polynomials:

(i) $x^2 + x$

(ii) $x - x^3$

(iii) $y + y^2 + 4$

(iv) $1 + x$

(v) 3t

(vi) $r^2$

(vii) $7x^3$

Solution:

Polynomials are classified based on their degree:

Linear polynomial: Degree 1
Quadratic polynomial: Degree 2
Cubic polynomial: Degree 3

Let's classify each polynomial:

(i) $x^2 + x$ : The highest power of $x$ is 2. So, it is a quadratic polynomial.

(ii) $x - x^3$ : The highest power of $x$ is 3. So, it is a cubic polynomial.

(iii) $y + y^2 + 4$ : The highest power of $y$ is 2. So, it is a quadratic polynomial.

(iv) $1 + x$ : The highest power of $x$ is 1. So, it is a linear polynomial.

(v) 3t: The highest power of $t$ is 1. So, it is a linear polynomial.

(vi) $r^2$ : The highest power of $r$ is 2. So, it is a quadratic polynomial.

(vii) $7x^3$ : The highest power of $x$ is 3. So, it is a cubic polynomial.

Question 1

Find the value of the polynomial $5x - 4x^2 + 3$ at:

(i) $x = 0$

(ii) $x = -1$

(iii) $x = 2$

Solution:

Let the given polynomial be p(x) = 5x - 4x^2 + 3. We need to find the value of p(x) for the given values of x.

(i) At x = 0:

p(0) = 5(0) - 4(0)^2 + 3

p(0) = 0 - 4(0) + 3

p(0) = 0 - 0 + 3

p(0) = 3

(ii) At x = -1:

p(-1) = 5(-1) - 4(-1)^2 + 3

p(-1) = -5 - 4(1) + 3

p(-1) = -5 - 4 + 3

p(-1) = -9 + 3

p(-1) = -6

(iii) At x = 2:

p(2) = 5(2) - 4(2)^2 + 3

p(2) = 10 - 4(4) + 3

p(2) = 10 - 16 + 3

p(2) = -6 + 3

p(2) = -3

Question 2

Find $p(0)$ , $p(1)$ and $p(2)$ for each of the following polynomials:

(i) $p(y) = y^2 - y + 1$

(ii) $p(t) = 2 + t + 2t^2 - t^3$

(iii) $p(x) = x^3$

(iv) $p(x) = (x - 1)(x + 1)$

Solution:

(i) For p(y) = y^2 - y + 1:

p(0) = (0)^2 - (0) + 1 = 0 - 0 + 1 = 1

p(1) = (1)^2 - (1) + 1 = 1 - 1 + 1 = 1

p(2) = (2)^2 - (2) + 1 = 4 - 2 + 1 = 3

(ii) For p(t) = 2 + t + 2t^2 - t^3:

p(0) = 2 + (0) + 2(0)^2 - (0)^3 = 2 + 0 + 0 - 0 = 2

p(1) = 2 + (1) + 2(1)^2 - (1)^3 = 2 + 1 + 2(1) - 1 = 2 + 1 + 2 - 1 = 4

p(2) = 2 + (2) + 2(2)^2 - (2)^3 = 2 + 2 + 2(4) - 8 = 4 + 8 - 8 = 4

(iii) For p(x) = x^3:

p(0) = (0)^3 = 0

p(1) = (1)^3 = 1

p(2) = (2)^3 = 8

(iv) For p(x) = (x - 1)(x + 1):

p(0) = (0 - 1)(0 + 1) = (-1)(1) = -1

p(1) = (1 - 1)(1 + 1) = (0)(2) = 0

p(2) = (2 - 1)(2 + 1) = (1)(3) = 3

Question 3

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

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

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

(iii) $p(x) = x^2 - 1$ , $x = 1, -1$

(iv) $p(x) = (x + 1)(x - 2)$ , $x = -1, 2$

(v) $p(x) = x^2$ , $x = 0$

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

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

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

Solution:

To verify if a value is a zero of a polynomial, we substitute the value into the polynomial. If the result is 0, then it is a zero.

(i) For p(x) = 3x + 1 and x = -\frac{1}{3}:

p(-\frac{1}{3}) = 3(-\frac{1}{3}) + 1 = -1 + 1 = 0.

So, x = -\frac{1}{3} is a zero of p(x).

(ii) For p(x) = 5x - \pi and x = \frac{4}{5}:

p(\frac{4}{5}) = 5(\frac{4}{5}) - \pi = 4 - \pi.

Since 4 - \pi

eq 0, x = \frac{4}{5} is not a zero of p(x).

(iii) For p(x) = x^2 - 1 and x = 1, -1:

For x = 1: p(1) = (1)^2 - 1 = 1 - 1 = 0. So, x = 1 is a zero.

For x = -1: p(-1) = (-1)^2 - 1 = 1 - 1 = 0. So, x = -1 is a zero.

(iv) For p(x) = (x + 1)(x - 2) and x = -1, 2:

For x = -1: p(-1) = (-1 + 1)(-1 - 2) = (0)(-3) = 0. So, x = -1 is a zero.

For x = 2: p(2) = (2 + 1)(2 - 2) = (3)(0) = 0. So, x = 2 is a zero.

(v) For p(x) = x^2 and x = 0:

p(0) = (0)^2 = 0.

So, x = 0 is a zero of p(x).

(vi) For p(x) = lx + m and x = -\frac{m}{l}:

p(-\frac{m}{l}) = l(-\frac{m}{l}) + m = -m + m = 0.

So, x = -\frac{m}{l} is a zero of p(x).

(vii) For p(x) = 3x^2 - 1 and x = -\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}:

For x = -\frac{1}{\sqrt{3}}: p(-\frac{1}{\sqrt{3}}) = 3(-\frac{1}{\sqrt{3}})^2 - 1 = 3(\frac{1}{3}) - 1 = 1 - 1 = 0. So, x = -\frac{1}{\sqrt{3}} is a zero.

For x = \frac{2}{\sqrt{3}}: p(\frac{2}{\sqrt{3}}) = 3(\frac{2}{\sqrt{3}})^2 - 1 = 3(\frac{4}{3}) - 1 = 4 - 1 = 3.

Since 3

eq 0, x = \frac{2}{\sqrt{3}} is not a zero.

(viii) For p(x) = 2x + 1 and x = \frac{1}{2}:

p(\frac{1}{2}) = 2(\frac{1}{2}) + 1 = 1 + 1 = 2.

Since 2

eq 0, x = \frac{1}{2} is not a zero of p(x).

Common mistakes

Incorrectly identifying expressions with fractional or negative exponents as polynomials.
Confusing the coefficient of a term with the term itself.
Errors in calculation when evaluating polynomials, especially with negative numbers.
Misclassifying polynomials based on their degree.
Incorrectly verifying zeroes, often due to calculation errors.

Revision tips

Focus on the definition of a polynomial and the conditions for an expression to be a polynomial.
Practice identifying coefficients and degrees for various polynomial forms.
Work through the evaluation examples carefully, paying attention to signs and order of operations.
Use the verification of zeroes exercises to solidify the understanding of what a zero means.
Review the classification of polynomials (linear, quadratic, cubic) to quickly identify their nature.

Practice MCQs

Q1. Which of the following is a polynomial in one variable?

Q2. What is the coefficient of $x^2$ in the polynomial 2 - $x^2$ + $x^3$ ?

Q3. Which of the following is a binomial of degree 35?

Q4. What is the degree of the polynomial 5t - √7?

Q5. The polynomial y + $y^2$ + 4 is classified as:

Q6. If p(x) = 5x - 4 $x^2$ + 3, what is the value of p(0)?

Q7. For the polynomial p(t) = 2 + t + 2 $t^2$ - $t^3$ , what is p(1)?

Q8. Is x = -1/3 a zero of the polynomial p(x) = 3x + 1?

Frequently asked questions

What is a polynomial in one variable?

A polynomial in one variable is an algebraic expression consisting of variables and coefficients, where the exponents of the variable are non-negative integers. For example, 4x^2 - 3x + 7 is a polynomial in one variable, x.

How do I find the degree of a polynomial?

The degree of a polynomial is the highest exponent of the variable present in the polynomial. For example, in 5x^3 + 4x^2 + 7x, the highest exponent is 3, so the degree is 3.

What are linear, quadratic, and cubic polynomials?

These are classifications based on the degree: a linear polynomial has degree 1 (e.g., 1 + x), a quadratic polynomial has degree 2 (e.g., x^2 + x), and a cubic polynomial has degree 3 (e.g., x - x^3).

How do I evaluate a polynomial?

To evaluate a polynomial for a given value of the variable, substitute that value into the polynomial expression and simplify. For instance, to find p(2) for p(x) = 5x - 4x^2 + 3, substitute x=2: p(2) = 5(2) - 4(2)^2 + 3 = 10 - 16 + 3 = -3.

What does it mean to verify if a value is a zero of a polynomial?

A value is a zero of a polynomial if, when substituted into the polynomial, the result is zero. For example, to verify if x = -1/3 is a zero of p(x) = 3x + 1, calculate p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0. Since the result is 0, it is a zero.

Why is 3√t + t√2 not a polynomial?

It is not a polynomial because the term 3√t can be written as 3t^(1/2). The exponent 1/2 is not a whole number, which is a requirement for a polynomial.

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CBSE Class 9 Maths Chapter 2 Polynomials NCERT Solutions

Quick info

Chapter summary

Learning outcomes

Topics covered

Paper topics

Important topics

PDF preview

Questions and Solutions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 1

Question 2

Question 3

Common mistakes

Revision tips

Practice MCQs

Frequently asked questions

Important links