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

Jul 26, 2015

Factor ${x}^{3} - 2 {x}^{2} - 4 x + 8$

Ans: ${\left(x - 2\right)}^{2} \left(x + 2\right)$

#### Explanation:

Factor by grouping:

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

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

Jul 26, 2015

Factor by grouping first, then use the formula for the difference of two squares.

#### Explanation:

${x}^{3} - 2 {x}^{2} - 4 x + 8$

You can factor this expression by grouping the first and third terms

$\left({x}^{3} - 4 x\right) - 2 {x}^{2} + 8$

This is equivalent to

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

The first binomial is actually the difference of two perfect squares, which means that you can write it as

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

${x}^{3} - 2 {x}^{2} - 4 x + 8 = \textcolor{g r e e n}{\left(x - 2\right) \left(x + 2\right) \left(x - 2\right)}$