# How do you factor 2x^3+3x^2-16x-24?

May 25, 2015

Given $2 {x}^{3} + 3 {x}^{2} - 16 x - 24$

Regroup as
$\textcolor{red}{2 {x}^{3} - 16 x} + \textcolor{b l u e}{3 {x}^{2} - 24}$

Extract factor from each pair
= color(red)((2x)(x^2-8)) + color(blue)((3)(x^2-8)

Extract the $\left({x}^{2} - 8\right)$ common factor
$= \left(2 x + 3\right) \left({x}^{2} - 8\right)$

If you are willing to employ irrational constants, this can be further factored (using the difference of squares) as
$= \left(2 x - 3\right) \left(x + 2 \sqrt{2}\right) \left(x - 2 \sqrt{2}\right)$

May 25, 2015

Notice that the ratio between the 1st and 2nd term is the same as that between the 3rd and 4th term. So grouping will work...

$2 {x}^{3} + 3 {x}^{2} - 16 x - 24 = \left(2 {x}^{3} + 3 {x}^{2}\right) - \left(16 x + 24\right)$

$= {x}^{2} \left(2 x + 3\right) - 8 \left(2 x + 3\right)$

$= \left({x}^{2} - 8\right) \left(2 x + 3\right)$

$= \left(x - \sqrt{8}\right) \left(x + \sqrt{8}\right) \left(2 x + 3\right)$

$= \left(x - 2 \sqrt{2}\right) \left(x + 2 \sqrt{2}\right) \left(2 x + 3\right)$