Question

If one zero of the polynomial ax³ + 7x-6 is 2, find the value of a and other zeros.​Please answer step by step

Answer 1

$\large\underline{\sf{Solution-}}$

Given polynomial is

$\rm :\longmapsto\:p(x) = {ax}^{3} + 7x - 6$

It is given that, one zero of polynomial p(x) is 2.

By factor theorem,

$\rm :\longmapsto\:p(2) = 0$

$\rm :\longmapsto\:a {(2)}^{3} + 7 \times 2 - 6 = 0$

$\rm :\longmapsto\:8a + 14 - 6 = 0$

$\rm :\longmapsto\:8a + 8= 0$

$\rm :\longmapsto\:8a = - \: 8$

$\bf\implies \:a = - 1$

Thus, given polynomial p(x) can be rewritten as

$\rm :\longmapsto\:p(x) = { - x}^{3} + 7x - 6$

$\rm \: \: = \: - ( {x}^{3} - 7x + 6)$

$\rm \: \: = \: - ( {x}^{3} - x - 6x+ 6)$

$\rm \: \: = - \: \bigg(x( {x}^{2} - 1) - 6(x - 1) \bigg)$

$\rm \: \: = \: - \bigg(x(x - 1)(x + 1) - 6(x - 1) \bigg)$

$\rm \: \: = \: - (x - 1)\bigg(x(x + 1) - 6\bigg)$

$\rm \: \: = \: - (x - 1)\bigg( {x}^{2} + x - 6\bigg)$

$\rm \: \: = \: - (x - 1)\bigg( {x}^{2} + 3x - 2x - 6\bigg)$

$\rm \: \: = \: - (x - 1)\bigg(x(x + 3) - 2(x + 3)\bigg)$

$\rm \: \: = \: - (x - 1)(x + 3)(x - 2)$

$\bf\implies \:p(x) = \: - \: (x - 1)(x + 3)(x - 2)$

Hence,

$\bf :\longmapsto\:Zeroes \: of \: p(x) = 1, - 3, 2$

So,

• Other zeroes are 1 and - 3