# How do you solve the inequality x^3-x^2-6x>0?

Feb 15, 2015

Factor the expression ${x}^{3} - {x}^{2} - 6 x$ on the left side of the inequality and then evaluate for each term:

${x}^{3} - {x}^{2} - 6 x > 0$
$\rightarrow$ $\left(x\right) \left(x - 3\right) \left(x + 2\right) > 0$

Note that $x \ne 0$ since the left side must be $> 0$

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

if $x < 0$
then $\left(x - 3\right)$ will be negative
$\rightarrow$ $\left(x + 2\right)$ must be $> 0$
(so the product $\left(x\right) {\left(x - 3\right)}_{\neg} \left(x + 2\right)$ will be $> 0$
i.e (neg) $\times$ (neg) $\times$ (pos) )
$\rightarrow$ # (-2) < x < 0

Therefore
${x}^{3} - {x}^{2} - 6 x > 0$
for $x > 3$ or $\left(- 2\right) < x < 0$