# How do you solve x^3+x^2-16x-16=0?

##### 1 Answer
Sep 26, 2017

$x \in \left\{- 1 , - 4 , + 4\right\}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} {x}^{3} + {x}^{2} - 16 x - 16 = 0$

By observation we can see that $x = - 1$ is an obvious solution.
That is $\left(x + 1\right)$ is a factor of the left side of the given equation.

Using polynomial long division or synthetic division we get:
$\textcolor{w h i t e}{\text{XXX}} \left(x + 1\right) \left({x}^{2} - 16\right) = 0$

$\left({x}^{2} - 16\right)$ is obviously the difference of squares with
$\textcolor{w h i t e}{\text{XXX}} \left({x}^{2} - 16\right) = \left(x + 4\right) \left(x - 4\right)$

So we have
$\textcolor{w h i t e}{\text{XXX}} \left(x + 1\right) \left(x + 4\right) \left(x - 4\right) = 0$
which implies
$\left(x + 1\right) = 0 \textcolor{w h i t e}{\text{xxx")rarrcolor(white)("xxx}} x = - 1$
or
$\left(x + 4\right) = 0 \textcolor{w h i t e}{\text{xxx")rarrcolor(white)("xxx}} x = - 4$
or
$\left(x - 4\right) = 0 \textcolor{w h i t e}{\text{xxx")rarrcolor(white)("xxx}} x = 4$