Question

Find the value of $a$ so that $a+x-4x^2+x^3$ is exactly divisible by $x-3$

Answer 1

Given one root is $(x-3)$

$\Rightarrow (x-3)$ divides $f\left(x\right)$ exactly.

$\Rightarrow x-3=0$

$\Rightarrow x=3$

Let $f\left(x\right)=a+x-4x^2+x^3$

$\therefore f\left(3\right)=0$

$\Rightarrow f\left(3\right)=a+3-4\times 3^2+3^3=0$

$\Rightarrow a+3-36+27=0$

$\Rightarrow a-6=0$

$\therefore a=6$