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

Jun 1, 2018

$- 4 < x < - 1$ or $x > 4$

#### Explanation:

We have
${x}^{3} + {x}^{2} - 16 x - 16 = \left(x - 4\right) \left(x + 1\right) \left(x + 4\right)$
For
$x > 4$
is this product positive.

The product also positive for

$- 4 < x < - 1$

Jun 1, 2018

The solution is $x \in \left(- 4 , - 1\right) \cup \left(4 , + \infty\right)$

#### Explanation:

The inequality is

${x}^{3} + {x}^{2} - 16 x - 16 > 0$

Factorising,

$\left({x}^{2}\right) \left(x + 1 - 16 \left(x + 1\right)\right) > 0$

$\left({x}^{2} - 16\right) \left(x + 1\right) > 0$

$\left(x + 4\right) \left(x - 4\right) \left(x + 1\right) > 0$

Let $f \left(x\right) = \left(x + 4\right) \left(x - 4\right) \left(x + 1\right)$

Build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$4$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 4$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 4$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x\right) > 0$ when $x \in \left(- 4 , - 1\right) \cup \left(4 , + \infty\right)$

graph{x^3+x^2-16x-16 [-19.55, 26.05, -4.47, 18.34]}