# How do you factor x^9+1?

Aug 9, 2016

${x}^{9} + 1 = \left(x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{\pi}{9}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{3 \pi}{9}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{5 \pi}{9}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{7 \pi}{9}\right) x + 1\right)$

#### Explanation:

The easiest way to do this is probably using Complex arithmetic and de Moivre's formula:

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta + i \sin n \theta$

We find that the nine $9$th roots of $- 1$ are:

$\cos \left(\frac{k \pi}{9}\right) + i \sin \left(\frac{k \pi}{9}\right)$

for $k = \pm 1 , \pm 3 , \pm 5 , \pm 7 , 9$

In order to find factors with Real coefficients, we can pair up the Complex conjugate pairs like this:

$\left(x - \cos \left(\frac{k \pi}{9}\right) - i \sin \left(\frac{k \pi}{9}\right)\right) \left(x - \cos \left(\frac{- k \pi}{9}\right) - i \sin \left(\frac{- k \pi}{9}\right)\right)$

$= \left(x - \cos \left(\frac{k \pi}{9}\right) - i \sin \left(\frac{k \pi}{9}\right)\right) \left(x - \cos \left(\frac{k \pi}{9}\right) + i \sin \left(\frac{k \pi}{9}\right)\right)$

$= {\left(x - \cos \left(\frac{k \pi}{9}\right)\right)}^{2} - {\left(i \sin \left(\frac{k \pi}{9}\right)\right)}^{2}$

$= {x}^{2} - 2 \cos \left(\frac{k \pi}{9}\right) x + 1$

for $k = 1 , 3 , 5 , 7$

Hence:

${x}^{9} + 1 = \left(x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{\pi}{9}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{3 \pi}{9}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{5 \pi}{9}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{7 \pi}{9}\right) x + 1\right)$