# How do you factor x^3+3x^2-9x-27?

Feb 1, 2017

${x}^{3} + 3 {x}^{2} - 9 x - 27 = \left(x - 3\right) \left(x + 3\right) \left(x + 3\right)$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We will use this with $a = x$ and $b = 3$.

First note that the ratio between the first and second terms of the given cubic is the same as that between the third and fourth terms.

So this cubic will factor by grouping:

${x}^{3} + 3 {x}^{2} - 9 x - 27 = \left({x}^{3} + 3 {x}^{2}\right) - \left(9 x + 27\right)$

$\textcolor{w h i t e}{{x}^{3} + 3 {x}^{2} - 9 x - 27} = {x}^{2} \left(x + 3\right) - 9 \left(x + 3\right)$

$\textcolor{w h i t e}{{x}^{3} + 3 {x}^{2} - 9 x - 27} = \left({x}^{2} - 9\right) \left(x + 3\right)$

$\textcolor{w h i t e}{{x}^{3} + 3 {x}^{2} - 9 x - 27} = \left({x}^{2} - {3}^{2}\right) \left(x + 3\right)$

$\textcolor{w h i t e}{{x}^{3} + 3 {x}^{2} - 9 x - 27} = \left(x - 3\right) \left(x + 3\right) \left(x + 3\right)$