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.