# How do you factor x^3+2x^2-8x-16?

Aug 28, 2016

${x}^{3} + 2 {x}^{2} - 8 x - 16 = \left(x - 2 \sqrt{2}\right) \left(x + 2 \sqrt{2}\right) \left(x + 2\right)$

#### Explanation:

Given:

${x}^{3} + 2 {x}^{2} - 8 x - 16$

Notice that the ratio between the first and second terms is the same as that between the third and fourth terms.

So this cubic will factor by grouping:

${x}^{3} + 2 {x}^{2} - 8 x - 16$

$= \left({x}^{3} + 2 {x}^{2}\right) - \left(8 x + 16\right)$

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

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

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

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