Class 9 NCERT Mathematics Solutions Chapter 2 Polynomials Exercise 2.2 (original) (raw)

Last Updated : 23 Jul, 2025

Chapter 2 of the Class 9 NCERT Mathematics textbook, titled "Polynomials," covers the fundamentals of polynomials, including their definitions, operations, and properties. Exercise 2.2 focuses on solving problems related to polynomial expressions and equations.

This section provides detailed solutions for Exercise 2.2 from Chapter 2 of the Class 9 NCERT Mathematics textbook. The exercise involves problems that require applying polynomial concepts such as polynomial operations, factorization, and evaluating polynomial expressions. Solutions are presented step-by-step to aid in understanding and solving polynomial problems effectively.

Polynomials

Polynomials are algebraic expressions that consist of the variables raised to whole number exponents and their coefficients. They play a significant role in algebra and are foundational in solving equations, graphing functions, and understanding calculus. This chapter and exercise focus on simplifying polynomials performing the arithmetic operations and factoring them. Understanding these concepts is vital for students to progress in their mathematical studies and applications.

Class 9 NCERT Mathematics Solutions- Exercise 2.2

**Question 1: Find the value of the polynomial (x) = 5x − 4x 2 + 3

****(i) x = 0**

****(ii) x = –1**

****(iii) x = 2**

**Solution:

Given equation: 5x − 4x2 + 3

Therefore, let f(x) = 5x - 4x2 + 3

****(i)** When x = 0

f(0) = 5(0)-4(0)2+3

= 3

****(ii)** When x = -1

f(x) = 5x−4x2+3

f(−1) = 5(−1)−4(−1)2+3

= −5–4+3

= −6

****(iii)** When x = 2

f(x) = 5x−4x2+3

f(2) = 5(2)−4(2)2+3

= 10–16+3

= −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)** p(y) = y2 – y + 1

Given equation: p(y) = y2–y+1

Therefore, p(0) = (0)2−(0)+1 = 1

p(1) = (1)2–(1)+1 = 1

p(2) = (2)2–(2)+1 = 3

Hence, p(0) = 1, p(1) = 1, p(2) = 3, for the equation p(y) = y2–y+1

****(ii)** p(t) = 2 + t + 2t2 − t3

Given equation: p(t) = 2+t+2t2−t3

Therefore, p(0) = 2+0+2(0)2–(0)3 = 2

p(1) = 2+1+2(1)2–(1)3 = 2+1+2–1 = 4

p(2) = 2+2+2(2)2–(2)3 = 2+2+8–8 = 4

Hence, p(0) = 2,p(1) =4 , p(2) = 4, for the equation p(t)=2+t+2t2−t3

****(iii)** p(x) = x3

Given equation: p(x) = x3

Therefore, p(0) = (0)3 = 0

p(1) = (1)3 = 1

p(2) = (2)3 = 8

Hence, p(0) = 0,p(1) = 1, p(2) = 8, for the equation p(x)=x3

****(iv)** p(x) = (x−1)(x+1)

Given equation: p(x) = (x–1)(x+1)

Therefore, 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

Hence, p(0)= 1, p(1) = 0, p(2) = 3, for the equation p(x) = (x−1)(x+1)

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

****(i) p(x) = 3x+1, x=−1/3**

****(ii) p(x) = 5x–π, x = 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 = −m/l**

****(vii) p(x) = 3x** 2 −1, x = -1/√3 , 2/√3

****(viii) p(x) = 2x+1, x = 1/2**

**Solution:

****(i)** p(x)=3x+1, x=−1/3

Given: p(x)=3x+1 and x=−1/3

Therefore, substituting the value of x in equation p(x), we get.

For, x = -1/3

p(−1/3) = 3(-1/3)+1

= −1+1

= 0

Hence, p(x) of -1/3 = 0

****(ii)** p(x)=5x–π, x = 4/5

Given: p(x)=5x–π and x = 4/5

Therefore, substituting the value of x in equation p(x), we get.

For, x = 4/5

p(4/5) = 5(4/5)–π

= 4–π

Hence, p(x) of 4/5 ≠ 0

****(iii)** p(x)=x2−1, x=1, −1

Given: p(x)=x2−1 and x=1, −1

Therefore, substituting the value of x in equation p(x), we get.

For x = 1

p(1) = 12−1

=1−1

= 0

For, x = -1

p(−1) = (-1)2−1

= 1−1

= 0

Hence, p(x) of 1 and -1 = 0

****(iv)** p(x) = (x+1)(x–2), x =−1, 2

Given: p(x) = (x+1)(x–2) and x =−1, 2

Therefore, substituting the value of x in equation p(x), we get.

For, x = −1

p(−1) = (−1+1)(−1–2)

= (0)(−3)

= 0

For, x = 2

p(2) = (2+1)(2–2)

= (3)(0)

= 0

Hence, p(x) of −1, 2 = 0

****(v)** p(x) = x2, x = 0

Given: p(x) = x2 and x = 0

Therefore, substituting the value of x in equation p(x), we get.

For, x = 0

p(0) = 02 = 0

Hence, p(x) of 0 = 0

****(vi)** p(x) = lx+m, x = −m/l

Given: p(x) = lx+m and x = −m/l

Therefore, substituting the value of x in equation p(x), we get.

For, x = −m/l

p(-m/l)= l(-m/l)+m

= −m+m

= 0

Hence, p(x) of -m/l = 0

****(vii)** p(x) = 3x2−1, x = -1/√3 , 2/√3

Given: p(x) = 3x2−1 and x = -1/√3 , 2/√3

Therefore, substituting the value of x in equation p(x), we get.

For, x = -1/√3

p(-1/√3) = 3(-1/√3)2 -1

= 3(1/3)-1

= 1-1

= 0

For, x = 2/√3

p(2/√3) = 3(2/√3)2 -1

= 3(4/3)-1

= 4−1

=3 ≠ 0

Hence, p(x) of -1/√3 = 0

but, p(x) of 2/√3 ≠ 0

****(viii)** p(x) =2x+1, x = 1/2

Given: p(x) =2x+1 and x = 1/2

Therefore, substituting the value of x in equation p(x), we get.

For, x = 1/2

p(1/2) = 2(1/2)+1

= 1+1

= 2≠0

Hence, p(x) of 1/2 ≠ 0

**Question 4: Find the zero of the polynomials in each of the following cases:

****(i) p(x) = x+5**

****(ii) p(x) = x–5**

****(iii) p(x) = 2x+5**

****(iv) p(x) = 3x–2**

****(vii) p(x) = cx+d, c ≠ 0, c, d are real numbers.**

**Solution:

****(i)** p(x) = x+5

Given: p(x) = x+5

To find the zero, let p(x) = 0

p(x) = x+5

0 = x+5

x = −5

Therefore, the zero of the polynomial p(x) = x+5 is when x = -5

****(ii)** p(x) = x–5

Given: p(x) = x–5

p(x) = x−5

x−5 = 0

x = 5

Therefore, the zero of the polynomial p(x) = x–5 is when x = 5

****(iii)** p(x) = 2x+5

Given: p(x) = 2x+5

p(x) = 2x+5

2x+5 = 0

2x = −5

x = -5/2

Therefore, the zero of the polynomial p(x) = 2x+5 is when x = -5/2

****(iv)** p(x) = 3x–2

Given: p(x) = 3x–2

p(x) = 3x–2

3x−2 = 0

3x = 2

x = 2/3

Therefore, the zero of the polynomial p(x) = 3x–2 is when x = 2/3

****(v)** p(x) = 3x

Given: p(x) = 3x

p(x) = 3x

3x = 0

x = 0

Therefore, the zero of the polynomial p(x) = 3x is when x = 0

****(vi)** p(x) = ax, a0

Given: p(x) = ax, a≠ 0

p(x) = ax

ax = 0

x = 0

Therefore, the zero of the polynomial p(x) = ax is when x = 0

****(vii)** p(x) = cx+d, c ≠ 0, c, d are real numbers.

Given: p(x) = cx+d

p(x) = cx + d

cx+d =0

x = -d/c

Therefore, the zero of the polynomial p(x) = cx+d is when x = -d/c

Conclusion

Chapter 2 of the Class 9 NCERT Mathematics textbook, "Polynomials," covers key concepts including polynomial operations, factorization, and evaluation. Exercise 2.2 involves applying these concepts to solve problems related to polynomial expressions. Detailed solutions are provided to help students understand and solve polynomial problems effectively.