# How do you completely factor x^3+2x-4x-8?

Jun 16, 2018

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

#### Explanation:

I assume that the correct equation is ${x}^{3} + \textcolor{red}{2 {x}^{2}} - 4 x - 8 = 0$

Now group the first two terms and the last two terms

$\left({x}^{3} + 2 {x}^{2}\right) + \left(- 4 x - 80\right) = 0$

Take out the common terms from each group

${x}^{2}$ from first group

$- 4$ from last group

${x}^{2} \left(x + 2\right) - 4 \left(x + 2\right) = 0$

Since $x + 2$ is common in both take it out

$\left(x + 2\right) \left({x}^{2} - 4\right) = 0$

${x}^{2} - 4$ can be factorize into $\left(x + 2\right) \left(x - 2\right)$

Then the factorized form of the equation will be

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

$\implies {\left(x + 2\right)}^{2} \left(x - 2\right) = 0$