# How do you factor 3x^8+45x^5+129x^2?

Jan 16, 2017

$3 {x}^{8} + 45 {x}^{5} + 129 {x}^{2}$

$= 3 {x}^{2} \left(x + \sqrt[3]{\frac{15}{2} - \frac{\sqrt{53}}{2}}\right) \left({x}^{2} - \left(\sqrt[3]{\frac{15}{2} - \frac{\sqrt{53}}{2}}\right) x + \sqrt[3]{\frac{139}{2} - \frac{15 \sqrt{53}}{2}}\right) \left(x + \sqrt[3]{\frac{15}{2} + \frac{\sqrt{53}}{2}}\right) \left({x}^{2} - \left(\sqrt[3]{\frac{15}{2} + \frac{\sqrt{53}}{2}}\right) x + \sqrt[3]{\frac{139}{2} + \frac{15 \sqrt{53}}{2}}\right)$

#### Explanation:

First note that all of the terms are divisible by $3 {x}^{2}$, so we can separate that out as a factor:

$3 {x}^{8} + 45 {x}^{5} + 129 {x}^{2} = 3 {x}^{2} \left({x}^{6} + 15 {x}^{3} + 43\right)$

We can treat the remaining sextic as a quadratic in ${x}^{3}$ to factor it:

${x}^{6} + 15 {x}^{3} + 43 = {\left({x}^{3} + \frac{15}{2}\right)}^{2} - {\left(\frac{15}{2}\right)}^{2} + 43$

$\textcolor{w h i t e}{{x}^{6} + 15 {x}^{3} + 43} = {\left({x}^{3} + \frac{15}{2}\right)}^{2} - \frac{225 - 172}{4}$

$\textcolor{w h i t e}{{x}^{6} + 15 {x}^{3} + 43} = {\left({x}^{3} + \frac{15}{2}\right)}^{2} - \frac{53}{4}$

$\textcolor{w h i t e}{{x}^{6} + 15 {x}^{3} + 43} = {\left({x}^{3} + \frac{15}{2}\right)}^{2} - {\left(\frac{\sqrt{53}}{2}\right)}^{2}$

$\textcolor{w h i t e}{{x}^{6} + 15 {x}^{3} + 43} = \left({x}^{3} + \frac{15}{2} - \sqrt{\frac{53}{2}}\right) \left({x}^{3} + \frac{15}{2} + \frac{\sqrt{53}}{2}\right)$

The sum of cubes identity can be written:

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

Note that:

${\left(\frac{15}{2} - \frac{\sqrt{53}}{2}\right)}^{2} = \frac{225}{4} - \frac{15 \sqrt{53}}{2} + \frac{53}{4} = \frac{139}{2} - \frac{15 \sqrt{53}}{2}$

${\left(\frac{15}{2} + \frac{\sqrt{53}}{2}\right)}^{2} = \frac{225}{4} + \frac{15 \sqrt{53}}{2} + \frac{53}{4} = \frac{139}{2} + \frac{15 \sqrt{53}}{2}$

Hence we find:

${x}^{3} + \frac{15}{2} - \frac{\sqrt{53}}{2} = {x}^{3} + {\left(\sqrt[3]{\frac{15}{2} - \frac{\sqrt{53}}{2}}\right)}^{3}$

$\textcolor{w h i t e}{{x}^{3} + \frac{15}{2} - \frac{\sqrt{53}}{2}} = \left(x + \sqrt[3]{\frac{15}{2} - \frac{\sqrt{53}}{2}}\right) \left({x}^{2} - \left(\sqrt[3]{\frac{15}{2} - \frac{\sqrt{53}}{2}}\right) x + \sqrt[3]{\frac{139}{2} - \frac{15 \sqrt{53}}{2}}\right)$

${x}^{3} + \frac{15}{2} + \frac{\sqrt{53}}{2} = {x}^{3} + {\left(\sqrt[3]{\frac{15}{2} + \frac{\sqrt{53}}{2}}\right)}^{3}$

$\textcolor{w h i t e}{{x}^{3} + \frac{15}{2} + \frac{\sqrt{53}}{2}} = \left(x + \sqrt[3]{\frac{15}{2} + \frac{\sqrt{53}}{2}}\right) \left({x}^{2} - \left(\sqrt[3]{\frac{15}{2} + \frac{\sqrt{53}}{2}}\right) x + \sqrt[3]{\frac{139}{2} + \frac{15 \sqrt{53}}{2}}\right)$

So putting it all together, we have:

$3 {x}^{8} + 45 {x}^{5} + 129 {x}^{2}$

$= 3 {x}^{2} \left(x + \sqrt[3]{\frac{15}{2} - \frac{\sqrt{53}}{2}}\right) \left({x}^{2} - \left(\sqrt[3]{\frac{15}{2} - \frac{\sqrt{53}}{2}}\right) x + \sqrt[3]{\frac{139}{2} - \frac{15 \sqrt{53}}{2}}\right) \left(x + \sqrt[3]{\frac{15}{2} + \frac{\sqrt{53}}{2}}\right) \left({x}^{2} - \left(\sqrt[3]{\frac{15}{2} + \frac{\sqrt{53}}{2}}\right) x + \sqrt[3]{\frac{139}{2} + \frac{15 \sqrt{53}}{2}}\right)$