# How do you determine the binomial factors of x^3-2x^2-4x+8?

Mar 3, 2018

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

#### Explanation:

Given:

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

Note that the ratio between the first and second terms is the same as that between the third and fourth terms.

So this quadrinomial will factor by grouping:

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

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

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

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

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

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