Question

Factorise: $x^3-2x^2-5x+6$

Answer 1

Let $f\left(x\right)=x^3-2x^2-5x+6$

The factors of the constant terms (i.e $+6$) are $\pm 1,\pm 2,\pm 3,\pm 6$

On substituting $x=1$ in $f\left(x\right),$ we get

$f\left(1\right)=(1{)}^3-2(1{)}^2-5\left(1\right)+6$

$f\left(1\right)=1-2-5+6=0$

When $f\left(1\right)=0,$ then $(x-1)$ is a factor of $f\left(x\right)$

Now divide the polynomial $f\left(x\right)$ by $(x-1)$

On dividing we get, $f\left(x\right)=(x-1)(x^2-x-6)$

Now $x^2-x-6=x^2-3x+2x-6$

$=x(x-3)+2(x-3)$

$=(x-3)(x+2)$

So, the other two roots are $x=3,$ and $x=-2$

Hence, $(x^3-2x^2-5x+6)=(x-1)(x-3)(x+2)$