If one zero of the polynomial ax³ + 7x-6 is 2, find the value of a an

1.	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» <P> $\large\underline{\sf{Solution-}}$ 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