# How do you solve m^3-2m^2-15m>0 using a sign chart?

##### 1 Answer
Mar 6, 2017

The solution is m in ]-3,0 [uu]5, +oo[

#### Explanation:

Let's factorise the inequality

${m}^{3} - 2 {m}^{2} - 15 m > 0$

$m \left({m}^{2} - 2 m - 15\right) > 0$

$m \left(m + 3\right) \left(m - 5\right) > 0$

Let $f \left(m\right) = m \left(m + 3\right) \left(m - 5\right)$

Now, we can build the sign chart

$\textcolor{w h i t e}{a a a a}$$m$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 3$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a a}$$5$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$m + 3$$\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}$$+$

$\textcolor{w h i t e}{a a a a}$$m$$\textcolor{w h i t e}{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}$$+$

$\textcolor{w h i t e}{a a a a}$$m - 5$$\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}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(m\right)$$\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}$$+$

Therefore,

$f \left(m\right) > 0$, when m in ]-3,0 [uu]5, +oo[