Question

If $3x-1$ is a factor of the polynomial $81x^3-45x^2+3a-6$, then the value of a is ______

Answer 1

Given: $81x^3-45x^2+3a-6$

Let, $f\left(x\right)=81x^3-45x^2+3a-6$

We know that if $(x+a)$ is the factor of the polynomial $f\left(x\right)$ then by factor theorem $f(-a)=0$

$\because (3x-1)$ is a factor of $f\left(x\right)$

$\therefore f\left(\frac{1}{3}\right)=0$

$\Rightarrow 81\times {\left(\frac{1}{3}\right)}^3-45\times {\left(\frac{1}{3}\right)}^2+3a-6=0$

$\Rightarrow \frac{81}{27}-\frac{45}{9}+3a-6=0$

$\Rightarrow 3a=8$

$\therefore a=\frac{8}{3}$