# JAC Class 9 Maths Solutions Chapter 2 Polynomials Ex 2.2

Jharkhand Board JAC Class 9 Maths Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.2

Page – 34

Question 1.
Find the value of the polynomial 5x + 4x2 + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2 ,
(i) p(x) = 5x + 4x2 + 3
p(0) = 5(0) + 4(0)2 + 3 = 3

(ii) p(x) = 5x + 4x2 + 3
P(-1) = 5(-1) + 4(-1)2 + 3
= -5 + 4(1) + 3 = 2

(iii) p(x) = 5x + 4x2 + 3
p(2) = 5(2) + 4(2)2 + 3
= 10 + 16 + 3 = 29

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y2 – y + 1
(ii) p(x) = x3
(iii) p(t) = 2 + t + 2t2 – t3
(iv) p(x) = (x – 1) (x + 1)
(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 = 3

(ii) p(x) = x3
P(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8

(iii) p(t) = 2 + t + 2t2 – t3
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

(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

Page – 35

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 = $$-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$$
(viii) p(x) = 2x + 1, x = $$\frac{1}{2}$$
(i) If x = $$– \frac{1}{3}$$ is a zero of polynomial p(x) then p($$– \frac{1}{3}$$) = 0
p($$– \frac{1}{3}$$) = 3($$– \frac{1}{3}$$) + 1 = -1 + 1 = 0
Therefore, x = $$– \frac{1}{3}$$ is a zero of polynomial p(x) = 3x + 1.

(ii) If x = $$\frac{4}{5}$$ is a zero of polynomial
p(x) = 5x – π then p($$\frac{4}{5}$$) should be 0.
p($$\frac{4}{5}$$) = 5($$\frac{4}{5}$$) – π = 4 – π ≠ 0
Therefore, x = $$\frac{4}{5}$$ is not a zero of given polynomial p(x) = 5x – π.

(iii) If x = 1 and x = – 1 are zeroes of polynomial p(x) = x2 – 1, then p(1) and p(-1) each should be 0.
p(1) = (1)2 – 1 = 0 and
P(-1) = (-1)2 – 1 = 0
Hence, x = 1 and – 1 are zeroes of the polynomial p(x) = x2 – 1.

(iv) If x = -1 and x = 2 are zeroes of polynomial p(x) = (x + 1) (x – 2), then p( -1) and p(2) each should be 0.
p(-1) = (-1 + 1) (-1 – 2) = 0 (-3) = 0, and
p(2) = (2 + 1) (2 – 2) = 3 (0) = 0
Therefore, x = – 1 and x = 2 are zeroes of the polynomial p(x) = (x+1) (x – 2).

(v) If x = 0 is a zero of polynomial p(x) = x2, then p(0) should be zero.
Here, p(0) = (0)2 = 0
Hence, x = 0 is a zero of the polynomial p(x) = x2.

(vi) If x = $$– \frac{m}{l}$$ is a zero of polynomial p(x) = lx + m, then p($$– \frac{m}{l}$$)= 0 should be 0.
p($$– \frac{m}{l}$$) = l($$– \frac{m}{l}$$) + m = -m + m = 0
Therefore x = ($$– \frac{m}{l}$$) is a zero of polynomial p(x) = lx + m

(vii) If x = $$-\frac{1}{\sqrt{3}}$$ and x = $$\frac{2}{\sqrt{3}}$$ are zeros of polynomial p(x) = 3x2 – 1, then p($$-\frac{1}{\sqrt{3}}$$) and p($$\frac{2}{\sqrt{3}}$$)should be 0.
p($$\frac{-1}{\sqrt{3}}$$) = $$3\left(\frac{-1}{\sqrt{3}}\right)^2$$ – 1 = 3($$\frac{1}{3}$$)-1 = 1 – 1 = 0
p($$\frac{2}{\sqrt{3}}$$) = $$3\left(\frac{2}{\sqrt{3}}\right)^2$$ – 1 = 3($$\frac{4}{3}$$) – 1 = 4 – 1 = 3 ≠ 0
Therefore, x = $$\frac{-1}{\sqrt{3}}$$ is a zero of polynomial p(x) = 3x2 – 1 but, x = $$\frac{2}{\sqrt{3}}$$ is not a zero of this polynomial.

(viii) If x = $$\frac{1}{2}$$ is a zero of polynomial p(x) = 2x + 1then p($$\frac{1}{2}$$) should be 0.
p($$\frac{1}{2}$$) = 2($$\frac{1}{2}$$) + 1 = 1 + 1 = 2 ≠ 0
Therefore, x = $$\frac{1}{2}$$ is not a zero of given polynomial p(x) = 2x + 1

Question 4.
Find the zero of the 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, d, are real numbers.
(i) p(x) = x + 5
p(x) = 0
x + 5 = 0
x = – 5
Therefore, x = – 5 is a zero of polynomial p(x) = x + 5 .

(ii) p(x) = x – 5
p(x) =0
x – 5 = 0
x = 5
Therefore, x = 5 is a zero of polynomial p(x) = x – 5.

(iii) p(x) = 2x + 5
p(x) = 0
2x + 5 = 0
2x = – 5
x = $$– \frac{5}{2}$$
Therefore, x = $$– \frac{5}{2}$$ is a zero of polynomial p(x) = 2x + 5.

(iv) p(x) = 3x – 2
p(x) = 0
3x – 2 = 0
x = $$\frac{2}{3}$$
Therefore, x = $$\frac{2}{3}$$ is a zero of polynomial p(x) = 3x – 2.

(v) p(x) = 3x
p(x) = 0
3x = 0
x = 0
Therefore, x = 0 is a zero of polynomial p(x) = 3x.

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

(vii) p(x) = cx + d, c ≠ 0
p(x) = 0
cx + d = 0
x = $$– \frac{d}{c}$$
Therefore, x = $$– \frac{d}{c}$$ is a zero of polynomial p(x) = cx + d.