# How do you solve (x-2)(x+5) >0?

May 17, 2015

The signs of the two factors change at $x = - 5$ and $x = 2$. So the sign of their product $\left(x - 2\right) \left(x + 5\right)$ will change at those two points.

In the case $x < - 5$:

$\left(x - 2\right) < 0$ and $\left(x + 5\right) < 0$ so $\left(x - 2\right) \left(x + 5\right) > 0$

In the case $x = - 5$:

$\left(x - 2\right) \left(x + 5\right) = 0$

In the case $- 5 < x < 2$:

$\left(x - 2\right) < 0$ but $\left(x + 5\right) > 0$ so $\left(x - 2\right) \left(x + 5\right) < 0$

In the case $x = 2$:

$\left(x - 2\right) \left(x + 5\right) = 0$

In the case $x > 2$:

$\left(x - 2\right) > 0$ and $\left(x + 5\right) > 0$ so $\left(x - 2\right) \left(x + 5\right) > 0$

Putting these cases together $\left(x - 2\right) \left(x + 5\right) > 0$ when $x < - 5$ or $x > 2$.