# How do solve x^3+2x^2<=8x and write the answer as a inequality and interval notation?

Oct 24, 2016

The answer is $- \infty \le x \le - 4$ and $0 \le x \le 2$

#### Explanation:

Let us factorise the inequality and make a sign chart

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

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

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

So the values are $x = 0$, $x = 2$and $x = - 4$

$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- 4$$\textcolor{w h i t e}{a a a a a}$$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$
$x$$\textcolor{w h i t e}{a a a a a}$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a 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}$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
$x + 4$$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a}$$-$$\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}$$+$
$X$$\textcolor{w h i t e}{a a a a a a}$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
So the answer is $- \infty \le x \le - 4$ and $0 \le x \le 2$