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

Feb 25, 2017

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

#### Explanation:

Note that the ratio of the first and second terms is the same as that of the third and fourth terms. So this cubic will factor by grouping:

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

$\textcolor{w h i t e}{{x}^{3} - 2 {x}^{2} - 4 x + 8} = {x}^{2} \left(x - 2\right) - 4 \left(x - 2\right)$

$\textcolor{w h i t e}{{x}^{3} - 2 {x}^{2} - 4 x + 8} = \left({x}^{2} - 4\right) \left(x - 2\right)$

$\textcolor{w h i t e}{{x}^{3} - 2 {x}^{2} - 4 x + 8} = \left({x}^{2} - {2}^{2}\right) \left(x - 2\right)$

$\textcolor{w h i t e}{{x}^{3} - 2 {x}^{2} - 4 x + 8} = \left(x - 2\right) \left(x + 2\right) \left(x - 2\right)$

$\textcolor{w h i t e}{{x}^{3} - 2 {x}^{2} - 4 x + 8} = {\left(x - 2\right)}^{2} \left(x + 2\right)$

Feb 26, 2017

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

#### Explanation:

Although this question has already been answered, here is an alternative way of grouping the 4 terms, which results in the same factors.

$\textcolor{b l u e}{{x}^{3}} \textcolor{red}{- 2 {x}^{2}} \textcolor{b l u e}{- 4 x} \textcolor{red}{+ 8}$

$= \textcolor{b l u e}{{x}^{3} - 4 x} \textcolor{red}{- 2 {x}^{2} + 8}$

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

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

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

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

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