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

Nov 23, 2016

The answer is $x \ge 2$ or x in [2,+oo[

#### Explanation:

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

Then $f \left(2\right) = 8 + 8 - 8 - 8 = 0$
Therefore, $\left(x - 2\right)$ is a factor of $f \left(x\right)$

$f \left(- 2\right) = - 8 + 8 + 8 - 8 = 0$
Therefore, $\left(x + 2\right)$ is a factor of $f \left(x\right)$

So, $\left(x + 2\right) \left(x - 2\right) = {x}^{2} - 4$ is a factor of $f \left(x\right)$

To find the last factor, let's do 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} - 4$

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

$\textcolor{w h i t e}{a a a a}$$0 + 2 {x}^{2}$$\textcolor{w h i t e}{a a a a a}$$0 - 8$

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

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

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

As ${\left(x + 2\right)}^{2} > 0$

Therefore, the sign of $f \left(x\right)$ will depend on the sign of $\left(x - 2\right)$

When $x < 2$, $f \left(x\right) < 0$

and when $x \ge 2$, $f \left(x\right) \ge 0$

graph{x^3+2x^2-4x-8 [-20.28, 20.27, -10.14, 10.14]}