Question

If $p\left(x\right)=x^3-4x^2+x+6$, then show that $p\left(3\right)=0$ and hence factorise $p\left(x\right)$

Answer 1

$P\left(x\right)=x^3-4x^2+x+6$

$P\left(3\right)=3^3-4\times 3^2+3+6=27-36+3+6=0$

$\therefore x-3$ is a factor of $x^3-4x^2+x+6$

$\therefore x^2-x-2$ is another factor.

$\therefore x^3-4x^2+x+6=(x-3)(x^2-x-2)$

$x^3-4x^2+x+6=(x-3)(x^2-2x+x-2)$

$x^3-4x^2+x+6=(x-3)(x(x-2)+(x-2))$

$x^3-4x^2+x+6=(x-3)(x-2)(x+1)$