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

##### 1 Answer
Nov 16, 2015

Factor by grouping and the sum of cubes identity to find:

${x}^{5} + {x}^{3} + 8 {x}^{2} + 8 = \left({x}^{2} + 1\right) \left(x + 2\right) \left({x}^{2} - 2 x + 4\right)$

#### Explanation:

The sum of cubes identity may be written:

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

We use this with $a = x$ and $b = 2$ to find:

${x}^{5} + {x}^{3} + 8 {x}^{2} + 8$

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

$= \left({x}^{2} + 1\right) {x}^{3} + \left({x}^{2} + 1\right) 8$

$= \left({x}^{2} + 1\right) \left({x}^{3} + 8\right)$

$= \left({x}^{2} + 1\right) \left({x}^{3} + {2}^{3}\right)$

$= \left({x}^{2} + 1\right) \left(x + 2\right) \left({x}^{2} - x \left(2\right) + {2}^{2}\right)$

$= \left({x}^{2} + 1\right) \left(x + 2\right) \left({x}^{2} - 2 x + 4\right)$