NCERT Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.2

NCERT Solutions Class 9 Maths
Chapter – 2 (Polynomials)

The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 2 Polynomials Exercise 2.2 has been provided here to help the students in solving the questions from this exercise.

Chapter 2: Polynomials

Exercise – 2.2

1. Find the value of the polynomial (x) = 5x − 4x²+ 3
(i) x = 0
(ii) x = – 1
(iii) x = 2
Answer –
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

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)

Answer –
(i) p(y) = y²− y + 1
p(0) = (0)²− (0) + 1
= 1

p(1)
= (1)²– (1) + 1
=1

p(2)
= (2)²– (2) + 1
= 4 – 2 +1
= 3

(ii) p(t) = 2 + t + 2t²− t³p(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

(iii) p(x) = x³p(0) = (0)³= 0

p(1) = (1)³= 1

p(2) = (2)³= 8

(iv) 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

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

Answer –
(i) p(x) = 3x + 1, x =−1/3
p(−1/3) = 3(-1/3) + 1
= −1 + 1
= 0
∴ – $\frac{1}{3}$ is a zero of p(x).

(ii) p(x) = 5x – π, x = 4/5
p(4/5) = 5 (4/5) – π
= 4 – π
∴ $\frac{4}{5}$ is not a zero of p(x).

(iii) p(x) = x²− 1, x = 1, −1
p(1) = 1²− 1
= 1 − 1
= 0

p(−1) = (-1)²− 1
= 1 − 1
= 0
∴1, −1 are zeros of p(x).

(iv) p(x) = (x + 1)(x – 2), x =−1, 2
p(−1) = (−1 + 1)(−1 – 2)
= (0)(−3)
= 0

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

(v) p(x) = x², x = 0
p(0) = 0²= 0
∴ 0 is a zero of p(x).

(vi) p(x) = lx + m, x = −m/l
p(-m/l) = l(-m/l) + m
= −m + m
= 0
∴ -m/l is a zero of p(x).

(vii) p(x) = 3x²−1, x = -1/√3 , 2/√3
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

∴ -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
p(1/2) = 2(1/2) + 1
= 1 + 1
= 2 ≠0
∴ 1/2 is not a zero of p(x).

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
(v) p (x) = 3x
(vi) p (x)= ax, a ≠ 0
(vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers.

Answer –
(i) p(x) = x+5
⇒ x + 5 = 0
⇒ x = −5
∴ -5 is a zero polynomial of the polynomial p(x).

(ii) p(x) = x–5
⇒ x−5 = 0
⇒ x = 5
∴ 5 is a zero polynomial of the polynomial p(x).

(iii) p(x) = 2x+5
⇒ 2x + 5 = 0
⇒ 2x = −5
⇒ x = -5/2
∴ x = -5/2 is a zero polynomial of the polynomial p(x).

(iv) p(x) = 3x–2
⇒ 3x − 2 = 0
⇒ 3x = 2
⇒ x = 2/3
∴ x = 2/3 is a zero polynomial of the polynomial p(x).

(v) p(x) = 3x
⇒ 3x = 0
⇒ x = 0
∴ 0 is a zero polynomial of the polynomial p(x).

(vi) p(x) = ax, a≠0
⇒ ax = 0
⇒ x = 0
∴ x = 0 is a zero polynomial of the polynomial p(x).

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
⇒ cx + d = 0
⇒ x = -d/c
∴ x = -d/c is a zero polynomial of the polynomial p(x).

NCERT Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.2