# How do you factor h(x)= x^3 + 2x^2 - 5x - 6?

Oct 23, 2015

Find $h \left(- 1\right) = 0$, then divide $h \left(x\right)$ by $\left(x + 1\right)$ to find the other factors.

#### Explanation:

First note that $h \left(- 1\right) = - 1 + 2 + 5 - 6 = 0$, so $\left(x + 1\right)$ is a factor of $h \left(x\right)$

Divide $h \left(x\right)$ by $\left(x + 1\right)$ using synthetic division:

So:

${x}^{3} + 2 {x}^{2} - 5 x - 6 = \left(x + 1\right) \left({x}^{2} + x - 6\right) = \left(x + 1\right) \left(x + 3\right) \left(x - 2\right)$