NCERT Solutions Class 9 Mathematics

Exercise 2.2

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

(i) x = 0 (ii) x = - 1 (iii) x = 2

Q.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)

Q.3. Verify whether the following are zeroes of the polynomial, indicated against them.

 (i) p(x) = 3x + 1, x = - (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 = - $\frac{\mathrm{m}}{\mathrm{l}}$ (vii) p(x) = 3x2 - 1, x = - $\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$ (viii) p(x) = 2x + 1, x = $\frac{1}{2}$

Q.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, c, d are real numbers.

NCERT Solutions Class 9 Mathematics

Exercise 2.2

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

(i) x = 0 (ii) x = - 1 (iii) x = 2

Solution:

(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

Q.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

We can calculate,

p(0) = (0)2 - (0) + 1 = 1

p(1) = (1)2 - (1) + 1 = 1

p(2) = (2)2 - (2) + 1 = 3

(ii) p(t) = 2 + t + 2t2 - t3

We can calculate,

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

(iii) p(x) = x3

We can calculate,

p(0) = (0)3 = 0

p(1) = (1)3 = 1

p(2) = (2)3 = 8

(iv) p(x) = (x - 1) (x + 1)

We can calculate,

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

Q.3. Verify whether the following are zeroes of the polynomial, indicated against them.

 (i) p(x) = 3x + 1, x = - (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 = - $\frac{\mathrm{m}}{\mathrm{l}}$ (vii) p(x) = 3x2 - 1, x = - $\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$ (viii) p(x) = 2x + 1, x = $\frac{1}{2}$

Solution:

If ‘a’ is a zero of polynomial p(x), then p(a) should be equal to 0.

(i) p(x) = 3x + 1; x = -

p$\left(-\frac{1}{3}\right)$ = 3$×\left(-\frac{1}{3}\right)$ + 1 = -1 + 1 = 0

Therefore, - $\frac{1}{3}$ is a zero of polynomial 3x + 1.

p$\left(\frac{4}{5}\right)$ = 5$\left(\frac{4}{5}\right)$ - π

= 4 - π ≠ 0

Therefore, $\frac{4}{5}$ is not a zero of polynomial 5x - π.

(iii) p(x) = x2 - 1; x = 1 and x = -1

p(1) = (1)2 - 1 = 0

p(-1) = (-1)2 - 1 = 0

Therefore, both 1 and -1 are zeroes of the polynomial x2 - 1.

(iv) p(x) = (x +1) (x - 2); x = -1, x = 2

p(-1) = (-1 + 1) (-1 - 2) = 0 (-3) = 0

p(2) = (2 + 1) (2 - 2) = 3 (0) = 0

Therefore, -1 and 2 are zeroes of the polynomial (x +1) (x - 2).

(v) p(x) = x2, x = 0

p(0) = (0)2 = 0

Therefore, 0 is a zero of the polynomial x2.

= - m + m = 0

Hence, $-\frac{\mathrm{m}}{\mathrm{l}}$ is a zero of the given polynomial.

= 1 - 1 =0

(viii)  p(x) = 2x + 1, x = $\frac{1}{2}$

= 1 + 1 = 2 ≠ 0

Therefore, is not a zero of given polynomial.

Q.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, c, d are real numbers.

Solution:

If ‘a’ is a zero of polynomial p(x), then p(a) should be equal to 0. So for finding the zeroes, we need to put p(x) = 0 and solve for x.

(i) p(x) = x + 5

x + 5 = 0

x = -5

Therefore, x = -5 is a zero of polynomial p(x) = x + 5

(ii) p(x) = x - 5

x - 5 = 0

x = 5

Therefore, x = 5 is a zero of polynomial p(x) = x - 5.

(iii) p(x) = 2x + 5

⇒ 2x + 5 = 0

⇒ 2x = - 5

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

(iv) p(x) = 3x - 2

⇒ 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

⇒ 3x = 0

x = 0

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

(vi) p(x) = ax

ax = 0

x = 0

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

(vii) p(x) = cx + d

cx + d = 0

Therefore, x = -  $\frac{\mathrm{d}}{\mathrm{c}}$ is a zero of polynomial p(x) = cx + d.