NCERT Solutions for Class 9 Maths Polynomials: Exercise 2

Question 1: Find the value of the polynomial (x) = 5x − 4x² + 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) = y²−y+1

(ii) p(t) = 2+t+2t²−t³

(iii) p(x) = x³

(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) = x²−1, x=1, −1

(iv) p(x) = (x+1)(x–2), x =−1, 2

(v) p(x) = x², x = 0

(vi) p(x) = lx+m, x = −m/l

(vii) p(x) = 3x²−1, 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

NCERT Solutions Class 9 Maths Chapter 2 Polynomials is curated by a team of professionals at GFG to help students in their academic journey. In this article, we have provided the answers to all the questions in the NCERT textbook. All of the problems in this chapter’s exercise from the NCERT textbook for Class 9 are covered in the NCERT Solutions for Class 9 Maths.

The topics covered in the Polynomials chapter in Class 9 are Polynomials, Types of polynomials, Operations on polynomials, Degree of a polynomial, Factorization of polynomials, Factor Theorem, Remainder theorem, Algebraic identities and Application problems are the important Chapters in Class 9 Maths.