Question

Factorise the following polynomial using synthetic division:
$2x^3-3x^2-3x+2$

Answer 1

$p\left(x\right)=2x^3-3x^2-3x+2$
$p\left(1\right)=2(1{)}^3-3(1{)}^2-3\left(1\right)+2$
$=2-3-3+2$
$=2-6$
$=-4$
$\ne 0$
$x-1$ is not a factor
$p(-1)=2(-1{)}^3-3(-1{)}^2-3(-1)+2$
$=-2-3+3+2$
$=5-5$
$=0$
$\therefore x+1$ is a factor

$$\begin{array}{cc}& 2x^2-5x+2=2x^2-4x-x+2\\ & =2x(x-2)-1(x-2)\\ & =(x-2)(2x-1)\end{array}$$
$\therefore$ The factors of $2x^3-3x^2-3x+2=(x+1)(x-2)(2x-1)$