# NCERT Solutions For Class 9 Math | Exercise 2.2

###### Chapter 2 | Polynomials & Zeroes of Polynomials | 4 Questions

Detailed solution to all exercise questions of NCERT book for CBSE Class 9 Math in Chapter 2 - Polynomials - Exercise 2.2. The exercise comprises 4 questions.

Key concepts covered in this exercise include value of polynomials for different values of variables, verifying whether the given values of variables are zeroes of polynomials, and finding the zeroes of polynomials.

1. Question 1

Find the value of the polynomial 5x - 4x2 + 3 at
(i) x = 0
(ii) x = - 1
(iii) x = 2

##### Solution

(i) x = 0
Substitute x = 0 in the polynomial 5x - 4x2 + 3
5(0) – 4(0)2 + 3 = 3

(ii) x = - 1
Substitute x = -1 in the polynomial 5x - 4x2 + 3
5(-1) – 4(- 1)2 + 3
- 5 - 4 + 3 = - 6

(iii) x = 2
Substitute x = 2 in the polynomial 5x - 4x2 + 3
5(2) – 4(2)2 + 3
10 – 16 + 3 = - 3

2. 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)

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 – π, x = $$frac{4}{5}$$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 = -$$frac{m}{l}$$vii) p(x) = 3x2 - 1, x = -$$frac{1}{√3}$, $\frac{2}{√3}$$viii) p(x) = 2x + 1, x = $$frac{1}{2}$ 4. Question 4 Find the zero of polynomial 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, c, d are real numbers

##### Solution

(i) p(x) = x + 5
The zero of the polynomial is the value of x for which p(x) = 0.
p(x) = x + 5 = 0
x = - 5
The zero of the polynomial p(x) = x + 5 is - 5

(ii) p(x) = x – 5
The zero of the polynomial is the value of x for which p(x) = 0.
p(x) = x – 5 = 0
x = 5
The zero of the polynomial p(x) = x – 5 is 5

(iii) p(x) = 2x + 5
The zero of the polynomial is the value of x for which p(x) = 0.
p(x) = 2x + 5 = 0
x = -$$frac{5}{2}$ The zero of the polynomial p$x) = 2x + 5 is -$$frac{5}{2}$$iv) p(x) = 3x – 2
The zero of the polynomial is the value of x for which p(x) = 0.
p(x) = 3x – 2 = 0
x = $$frac{2}{3}$ The zero of the polynomial p$x) = 3x – 2 is $$frac{2}{3}$$v) p(x) = 3x
The zero of the polynomial is the value of x for which p(x) = 0.
p(x) = 3x = 0
x = 0
The zero of the polynomial p(x) = 3x is 0

(vi) p(x) = ax, a ≠ 0
The zero of the polynomial is the value of x for which p(x) = 0.
p(x) = ax = 0
x = 0
The zero of the polynomial p(x) = ax is 0

(vii) p(x) = cx + d
The zero of the polynomial is the value of x for which p(x) = 0.
p(x) = cx + d = 0
x = -$$frac{d}{c}$ The zero of the polynomial p$x) = cx + d is -$\frac{d}{c}$

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