# How do you factor the expression 2x^4 + 16x?

Mar 10, 2016

$2 {x}^{4} + 16 x = 2 x \left(x + 2\right) \left({x}^{2} - 2 x + 4\right)$

#### Explanation:

The idea is to transform the expression in a form where you can use the remarkable factorization identities.

Here we can use Factoring by Grouping to obtain:

$2 {x}^{4} + 16 x = \textcolor{g r e e n}{2 x} \left({x}^{3} + 8\right) = 2 x \left({x}^{3} + {2}^{3}\right)$

Now, the second factor is a sum of cubes, then we can use the following identity:

$\left({a}^{3} + {b}^{3}\right) = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

where:

$a = x , b = 2$

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

No more factorization allowed