# How do you solve x^3+2x^2<=8x using a sign chart?

##### 1 Answer
Feb 21, 2017

The solution is x in ]-oo,-4]uu[0, 2]

#### Explanation:

Let's rewrite and factorise the inequality

${x}^{3} + 2 {x}^{2} \le 8 x$

${x}^{3} + 2 {x}^{2} - 8 x \le 0$

$x \left({x}^{2} + 2 x - 8\right) \le 0$

$x \left(x - 2\right) \left(x + 4\right) \le 0$

Let $f \left(x\right) = x \left(x - 2\right) \left(x + 4\right)$

Now, we can build the sign chart

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

$\textcolor{w h i t e}{a a a a}$$x - 2$$\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,

$f \left(x\right) \le 0$, when x in ]-oo,-4]uu[0, 2]