# How do you factor 2x^5+4x^4-4x^3-8x^2 ?

Mar 2, 2017

$2 {x}^{5} + 4 {x}^{4} - 4 {x}^{3} - 8 {x}^{2} = 2 {x}^{2} \left(x - \sqrt{2}\right) \left(x + \sqrt{2}\right) \left(x + 2\right)$

#### Explanation:

Notice 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.

Note that all of the terms are divisible by $2 {x}^{2}$ so that can be separated out first.

Finally we can use the difference of squares identity:

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

with $a = x$ and $b = \sqrt{2}$

$2 {x}^{5} + 4 {x}^{4} - 4 {x}^{3} - 8 {x}^{2} = 2 {x}^{2} \left({x}^{3} + 2 {x}^{2} - 2 x - 4\right)$

$\textcolor{w h i t e}{2 {x}^{5} + 4 {x}^{4} - 4 {x}^{3} - 8 {x}^{2}} = 2 {x}^{2} \left(\left({x}^{3} + 2 {x}^{2}\right) - \left(2 x + 4\right)\right)$

$\textcolor{w h i t e}{2 {x}^{5} + 4 {x}^{4} - 4 {x}^{3} - 8 {x}^{2}} = 2 {x}^{2} \left({x}^{2} \left(x + 2\right) - 2 \left(x + 2\right)\right)$

$\textcolor{w h i t e}{2 {x}^{5} + 4 {x}^{4} - 4 {x}^{3} - 8 {x}^{2}} = 2 {x}^{2} \left({x}^{2} - 2\right) \left(x + 2\right)$

$\textcolor{w h i t e}{2 {x}^{5} + 4 {x}^{4} - 4 {x}^{3} - 8 {x}^{2}} = 2 {x}^{2} \left({x}^{2} - {\left(\sqrt{2}\right)}^{2}\right) \left(x + 2\right)$

$\textcolor{w h i t e}{2 {x}^{5} + 4 {x}^{4} - 4 {x}^{3} - 8 {x}^{2}} = 2 {x}^{2} \left(x - \sqrt{2}\right) \left(x + \sqrt{2}\right) \left(x + 2\right)$