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

Oct 24, 2016

The expression is $\ge 0$ for $x > 2$

#### Explanation:

let $f \left(x\right) = {x}^{3} + 2 {x}^{2} - 4 x - 8$
Let us start by factorising the expression
By trial and error, ·$\left(x - 2\right)$ is a factor as $f \left(2\right) = 0$

let us make the long division
${x}^{3} + 2 {x}^{2} - 4 x - 8$$\textcolor{w h i t e}{a a a}$∣$x - 2$
${x}^{3} - 2 {x}^{2}$$\textcolor{w h i t e}{a a a a a a a a a a a}$∣${x}^{2} + 4 x + 4$
$0 + 4 {x}^{2} - 4 x$
$\textcolor{w h i t e}{a a a a}$$4 {x}^{2} - 8 x$
$\textcolor{w h i t e}{a a a a a}$$0 + 4 x - 8$
$\textcolor{w h i t e}{a a a a a a a a a}$$4 x - 8$
$\textcolor{w h i t e}{a a a a a a a a a a}$$0 + 0$

so $f \left(x\right) = \left(x - 2\right) \left({x}^{2} + 4 x + 4\right) = \left(x - 2\right) {\left(x + 2\right)}^{2}$
$f \left(x\right) \ge 0$
Looking at the factorisation
${\left(x + 2\right)}^{2} \ge 0$ for all values of x∈RR
$\left(x - 2\right) \ge 0$ for $x \ge 2$
So the answer is $x \ge 2$