# How do you factor y= x^5 - 16x^3 + 8x^2 - 128 ?

Jan 6, 2016

Factor by grouping then using a couple of identities to find:

${x}^{5} - 16 {x}^{3} + 8 {x}^{2} - 128 = \left(x + 2\right) \left({x}^{2} - 2 x + 4\right) \left(x - 4\right) \left(x + 4\right)$

#### Explanation:

The sum of cubes identity can be written:

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

The difference of squares identity can be written:

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

Factor by grouping then use these two identities as follows:

${x}^{5} - 16 {x}^{3} + 8 {x}^{2} - 128$

$= \left({x}^{5} - 16 {x}^{3}\right) + \left(8 {x}^{2} - 128\right)$

$= {x}^{3} \left({x}^{2} - 16\right) + 8 \left({x}^{2} - 16\right)$

$= \left({x}^{3} + 8\right) \left({x}^{2} - 16\right)$

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

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

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