# How do you solve x^3+2x^2-4x-8 <=0?

Apr 17, 2017

The solution is $x \in \left(- \infty , 2\right]$

#### Explanation:

Let $f \left(x\right) = {x}^{3} + 2 {x}^{2} - 4 x - 8$

$f \left(2\right) = 8 + 8 - 8 - 8 = 0$

Therefore,

$\left(x - 2\right)$ is a factor of $f \left(x\right)$

To find the other factors, perform a long division

$\textcolor{w h i t e}{a a a a}$${x}^{3} + 2 {x}^{2} - 4 x - 8$$\textcolor{w h i t e}{a a a a}$$|$$x - 2$

$\textcolor{w h i t e}{a a a a}$${x}^{3} - 2 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a a}$$|$${x}^{2} + 4 x + 4$

$\textcolor{w h i t e}{a a a a a}$$0 + 4 {x}^{2} - 4 x$

$\textcolor{w h i t e}{a a a a a a a}$$+ 4 {x}^{2} - 8 x$

$\textcolor{w h i t e}{a a a a a a a a a a}$$0 + 4 x - 8$

$\textcolor{w h i t e}{a a a a a a a a a a a a}$$+ 4 x - 8$

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$+ 0 - 0$

So,

$f \left(x\right) = \left(x - 2\right) \left({x}^{2} + 4 x + 4\right) = \left(x - 2\right) {\left(x + 2\right)}^{2}$

We 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 a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 2$$\textcolor{w h i t e}{a a a a}$$+ 2$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$${\left(x + 2\right)}^{2}$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$\left(x - 2\right)$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a 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 a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$

Therefore,

$f \left(x\right) \le 0$ when $x \in \left(- \infty , 2\right]$

graph{x^3+2x^2-4x-8 [-21.2, 19.34, -12.25, 8.03]}