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

Nov 15, 2015

Factor by grouping, sum of cubes identity and difference of squares identity 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:

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

Hence:

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

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

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

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

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

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

This has no simpler factors with Real coefficients.

If you allow Complex coefficients then:

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

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i = \cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)$ is the primitive Complex cube root of unity.