NCERT Solutions Class 9th Maths Chapter – 2 Polynomials Exercise – 2.2

Question 1. Find the value of the polynomial 5x – 4x² + 3 at

(i) x = 0

Solution
1et p(x) = 5x – 4x² + 3
p(0) = 5(0) – 4(0)² + 3
= 0 – 0 + 3
= 3
Thus, the value of 5x – 4x² + 3 at x = 0 is 3.

(ii) x = – 1

Solution
p(-1) = 5(-1) – 4(-1)² + 3
= – 5 – 4 + 3
= -9 + 3
= -6
Thus, the value of 5x – 4x² + 3 at x = -1 is -6.

(iii) x = 2

Solution
p(2) = 5(2) – 4(2)² + 3
= 10 – 4(4) + 3
= 10 – 16 + 3
= -3
Thus, the value of 5x – 4x² + 3 at x = 2 is – 3.

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

(i) p(y) = y² – y +1

Solution
Given that p(y) = y² – 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) p(t) = 2 + t + 2t² – t³

Solution
Given that p(t) = 2 + t + 2t² – t³
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
= 4
p( 2) = 2 + 2 + 2(2)² – (2)³
= 2 + 2 + 8 – 8
= 4

(iii) P(x) = x³

Solution
Given that p(x) = x³
p(0) = (0)³ = 0
p(1) = (1)³ = 1
p(2) = (2)³ = 8

(iv) p(x) = (x – 1) (x + 1)

Solution
Given that 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 = –1/3

Solution
For, x = -1/3, p(x) = 3x + 1
p(−1/3) = 3(-1/3)+1
= −1+1
= 0
Thus, -1/3 is a zero of p(x).

(ii) p (x) = 5x – π, x = 4/5

Solution
For, x = 4/5, p(x) = 5x – π
p(4/5) = 5(4/5)- π
= 4 – π
Thus, 4/5 is not a zero of p(x).

(iii) p(x) = x² – 1, x = 1, – 1

Solution
For, x = 1, −1
p(x) = x² − 1
p(1) = 1² − 1 = 1 − 1 = 0
p(−1) = (-1)² − 1 = 1 − 1 = 0
Thus, 1, −1 are zeros of p(x).

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

Solution
For, x = −1,2
p(x) = (x + 1) (x – 2)
p(−1) = (−1 + 1) (−1 – 2)
= (0) (−3)
= 0
p(2) = (2 + 1) (2 – 2)
= (3) (0)
= 0
Thus, −1,2 are zeros of p(x).

(v) p(x) = x², x = 0

Solution
For, x = 0 p(x) = x²
p(0) = 0² = 0
Thus, 0 is a zero of p(x).

(vi) p(x) = lx + m, x = – m/l

Solution
For, x = -m/l
p(x) = lx + m
p(-m/l) = l(-m/l) + m
= -m + m 
= 0
Thus, -m/l is a zero of p(x).

(vii) P(x) = 3x² – 1, x = – 1/√3, 2/√3

Solution
For, x = -1/√3, 2/√3
p(x) = 3x² − 1
p(-1/√3) = 3(-1/√3)² – 1
= 3(1/3) – 1
= 1-1
= 0
p(2/√3 ) = 3(2/√3)² – 1
= 3(4/3) – 1
= 4 – 1
=3 ≠ 0
Thus, -1/√3 is a zero of p(x) but 2/√3 is not a zero of p(x).

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

Solution
For, x = 1/2
p(x) = 2x + 1
p(1/2) = 2(1/2) + 1
= 1 + 1
= 2 ≠ 0
Thus, 1/2 is not a zero of p(x).

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

(i) p(x) = x + 5

Solution
We have, p(x) = x + 5
Since, p(x) = 0
⇒ x + 5 = 0
⇒ x = -5
Thus, zero of x + 5 is -5

(ii) p(x) = x – 5

Solution
We have, p(x) = x – 5
Since, p(x) = 0
⇒ x – 5 = 0
⇒ x = -5
Thus, zero of x – 5 is 5

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

Solution
We have, p(x) = 2x + 5
Since, p(x) = 0
⇒ 2x + 5 = 0
⇒ 2x = -5
⇒ x = –5/2
Thus, zero of 2x + 5 is –5/2

(iv) p(x) = 3x – 2

Solution
We have, p(x) = 3x – 2
Since, p(x) = 0
⇒ 3x – 2 = 0
⇒ 3x = 2
⇒ x = 2/3
Thus, zero of 3x – 2 is 2/3

(v) p(x) = 3x

Solution
We have, p(x) = 3x
Since, p(x) = 0
⇒ 3x = 0
⇒ x = 0
Thus, zero of 3x is 0

(vi) p(x) = ax, a ≠ 0

Solution
We have, p(x) = ax, a ≠ 0
Since, p(x) = 0
=> ax = 0
=> x = 0
Thus, zero of ax is 0

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

Solution
We have, p(x) = cx + d
Since, p(x) = 0
⇒ cx + d = 0
⇒ cx = -d
⇒ x = –d/c
Thus, zero of cx + d is – d/c