# How do solve 2x^3+x^2>6x and write the answer as a inequality and interval notation?

Feb 23, 2017

The solutions are $- 2 < x < 0$ and $x > \frac{3}{2}$
Or in interval notation, x in ]-2,0[uu]3/2,+oo[

#### Explanation:

Let's rewrite and factorise the inequality

$2 {x}^{3} + {x}^{2} > 6 x$

$2 {x}^{3} + {x}^{2} - 6 x > 0$

$x \left(2 {x}^{2} + x - 6\right) > 0$

$x \left(2 x - 3\right) \left(x + 2\right) > 0$

Let $f \left(x\right) = x \left(2 x - 3\right) \left(x + 2\right)$

We can build the sign chart

$\textcolor{w h i t e}{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}$$- 2$$\textcolor{w h i t e}{a a a a a}$$0$$\textcolor{w h i t e}{a a a a a}$$\frac{3}{2}$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$

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

Therefore,

$f \left(x\right) > 0$ when $- 2 < x < 0$ and $x > \frac{3}{2}$

Or in interval notation, x in ]-2,0[uu]3/2,+oo[