# How do you factor x^12+8?

Feb 16, 2016

Use some identities to find the quadratic factors of ${x}^{12} + 8$

#### Explanation:

First notice that ${x}^{12} \ge 0$ for any Real value of $x$. So ${x}^{12} + 8 \ge 8 > 0$ for any Real $x$ and this polynomial has no linear factors with Real coefficients.

We can find all of the quadratic factors as follows:

The sum of cubes identity can be written:

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

Hence:

${x}^{12} + 8 = {\left({x}^{4}\right)}^{3} + {2}^{3}$

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

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

Notice that:

$\left({a}^{2} - k a b + {b}^{2}\right) \left({a}^{2} + k a b + {b}^{2}\right) = \left({a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4}\right)$

In particular, if $k = \sqrt{2}$ then we find:

$\left({a}^{2} - \sqrt{2} a b + {b}^{2}\right) \left({a}^{2} + \sqrt{2} a b + {b}^{2}\right) = {a}^{4} + {b}^{4}$

Hence (putting $a = x$ and $b = \sqrt[4]{2}$) we find:

$\left({x}^{4} + 2\right) = \left({x}^{2} - \sqrt[4]{8} x + \sqrt{2}\right) \left({x}^{2} + \sqrt[4]{8} x + \sqrt{2}\right)$

If instead we put $k = \sqrt{3}$ then we find:

$\left({a}^{2} - \sqrt{3} a b + {b}^{2}\right) \left({a}^{2} + \sqrt{3} a b + {b}^{2}\right) = {a}^{4} - {a}^{2} {b}^{2} + {b}^{4}$

So if we let $a = {x}^{2}$ and $b = \sqrt{2}$ then we find:

$\left({x}^{4} - \sqrt{6} {x}^{2} + 2\right) \left({x}^{4} + \sqrt{6} {x}^{2} + 2\right) = {x}^{8} - 2 {x}^{4} + 4$

Putting $k = \sqrt{2 + \sqrt{3}}$, $a = x$ and $b = \sqrt[4]{2}$ we find:

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

$= {x}^{4} - \sqrt{6} {x}^{2} + 2$

Putting $k = \sqrt{2 - \sqrt{3}}$, $a = x$ and $b = \sqrt[4]{2}$ we find:

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

$= {x}^{4} + \sqrt{6} {x}^{2} + 2$

Putting this all together, we find:

${x}^{12} + 8 = \left({x}^{2} - \sqrt[4]{8} x + \sqrt{2}\right) \left({x}^{2} + \sqrt[4]{8} x + \sqrt{2}\right) \left({x}^{2} - \sqrt{2 + \sqrt{3}} \sqrt[4]{2} x + \sqrt{2}\right) \left({x}^{2} + \sqrt{2 + \sqrt{3}} \sqrt[4]{2} x + \sqrt{2}\right) \left({x}^{2} - \sqrt{2 - \sqrt{3}} \sqrt[4]{2} x + \sqrt{2}\right) \left({x}^{2} + \sqrt{2 - \sqrt{3}} \sqrt[4]{2} x + \sqrt{2}\right)$