Question

Factorise the following polynomial using synthetic division:
$x^3-3x^2-10x+24$

Answer 1

$x^3-3x^2-10x+24$

Let $p\left(x\right)=x^3-3x^2-10x+24$

Sum of all the co-efficient of even powers with constant

$=-3+24=21$

Sum of co-efficients of odd powers $=1-10=-921\ne -9$

Hence $(x+1)$ is not a factor.

$p\left(2\right)=2^3-3\left(2^2\right)-10\times 2+24=8-12-20+24$

$=32-32=0$

$\therefore (x-2)$ is a factor.

Now we use synthetic division to find other factorThus $(x-2)(x+3)(x-4)$ are the factors.

$\therefore x^3-3x^2-10x+24=(x-2)(x+3)(x-4)$