# How do you factor x^7 - 16x^5 + 5x^4 - 80x^2?

Sep 17, 2016

${x}^{7} - 16 {x}^{5} + 5 {x}^{4} - 80 {x}^{2}$

$= {x}^{2} \left(x + \sqrt[3]{5}\right) \left({x}^{2} - \sqrt[3]{5} x + \sqrt[3]{25}\right) \left(x - 4\right) \left(x + 4\right)$

#### 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:

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

Given:

${x}^{7} - 16 {x}^{5} + 5 {x}^{4} - 80 {x}^{2}$

All of the terms are divisible by ${x}^{2}$, so that can be separated out as a factor.

Note also 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:

${x}^{7} - 16 {x}^{5} + 5 {x}^{4} - 80 {x}^{2}$

$= {x}^{2} \left(\left({x}^{5} - 16 {x}^{3}\right) + \left(5 {x}^{2} - 80\right)\right)$

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

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

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

$= {x}^{2} \left(x + \sqrt[3]{5}\right) \left({x}^{2} - \sqrt[3]{5} x + \sqrt[3]{25}\right) \left(x - 4\right) \left(x + 4\right)$