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

Sep 26, 2017

$\left({w}^{3} - 3 {w}^{2} - 9 w + 27\right) = \textcolor{b l u e}{\left(x - 3\right) \left(x - 3\right) \left(x + 3\right)}$

#### Explanation:

Writing ${w}^{3} - 3 {w}^{2} - 9 w + 27$
as
$\textcolor{w h i t e}{\text{XXX}} {w}^{3} - 3 {w}^{2} - {3}^{3} w + {3}^{3}$
makes it fairly clear that
$\textcolor{w h i t e}{\text{XXX}} w = 3$ is a zero of this expression.

That is $\left(w - 3\right)$ is a factor of $\left({w}^{3} - 3 {w}^{2} - 9 w + 27\right)$

If we divided $\left({w}^{3} - 3 {w}^{2} - 9 w + 27\right)$ by $\left(w - 3\right)$
(either by polynomial long division or synthetic division)
we find
$\textcolor{w h i t e}{\text{XXX}} \left({x}^{2} - 9\right)$ is also a factor

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

So a complete factoring is
$\textcolor{w h i t e}{\text{XXX}} \left({w}^{3} - 3 {w}^{2} - 9 w + 27\right) = \left(x - 3\right) \left(x - 3\right) \left(x + 3\right)$