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.