# How do you factor x^3+3x^2-2x-6?

Sep 9, 2016

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

#### Explanation:

We have: ${x}^{3} + 3 {x}^{2} - 2 x - 6$

$= \left({x}^{3} + 3 {x}^{2}\right) + \left(- 2 x - 6\right)$

Let's take ${x}^{2}$ in common in the first set of parentheses and $- 2$ in common in the second set of parentheses, respectively:

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

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

Then, let's take $x + 3$ in common:

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

If you allow surds to be used, then the second set of parentheses can be factorised further:

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