Question

Factorise: $2x^3-10x^2-4x+24$
using synthetic division.

• A. $2(x+2)(x-3)(x+4)$
• B. $(x+2)(x-3)(x-4)$
• C. $2(x+2)(x-3)(x-4)$
• D. $2(x-2)(x-3)(x-4)$

Answer 1

The correct option is C $2(x+2)(x-3)(x-4)$

$p\left(x\right)=(2x^3-10x^2-4x+48)$
Using trial and error method,
when
$x=-2p(-2)=2\times (-2{)}^3-10\times (-2{)}^2-4\times (-2)+24)=-16+40-8-48=0$
Hence, one factor = $(x+2)$

Using synthetic division,
$\begin{array}{ccccc}& x^3& x^2& x^1& x^0\\ & 2& -10& -4& 48\\ -2& \downarrow & -4& 28& -48\\ & 2& -14& 24& |-0\end{array}$

$\therefore p\left(x\right)=(x-2)(2x^2-14x+24)$

Now, using trial and error method,
when,
$x=3q\left(x\right)=(2x^2-14x+24)=2\times 3^2-14\times 3+24=0$
Hence, another factor is $(x-3)$

$\begin{array}{cccc}& x^2& x^1& x^0\\ & 2& -14& 24\\ 3& \downarrow & 6& -24\\ & 2& -8& |-0\end{array}$

$\therefore p\left(x\right)=(x+2)(x-3)(2x-8)=2(x+2)(x-3)(x-4)$