# How do you factor x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x +1?

May 11, 2016

${x}^{8} + {x}^{7} + {x}^{6} + {x}^{5} + {x}^{4} + {x}^{3} + {x}^{2} + x + 1$

$= \left({x}^{2} + x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{2 \pi}{9}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{4 \pi}{9}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{8 \pi}{9}\right) x + 1\right)$

#### Explanation:

Notice that if you multiply this polynomial by $\left(x - 1\right)$ the result is ${x}^{9} - 1$.

So the zeros of this polynomial are all the Complex $9$th roots of $1$ except $1$.

Using the difference of cubes identity a couple of times, we find:

${x}^{9} - 1 = {\left({x}^{3}\right)}^{3} - {1}^{3} = \left({x}^{3} - 1\right) \left({x}^{6} + {x}^{3} + 1\right) = \left(x - 1\right) \left({x}^{2} + x + 1\right) \left({x}^{6} + {x}^{3} + 1\right)$

Then we can divide both ends by $\left(x - 1\right)$ to get:

${x}^{8} + {x}^{7} + {x}^{6} + {x}^{5} + {x}^{4} + {x}^{3} + {x}^{2} + x + 1 = \left({x}^{2} + x + 1\right) \left({x}^{6} + {x}^{3} + 1\right)$

No surprise there.

Note that the zeros of ${x}^{2} + x + 1$ are the Complex $3$rd roots of $1$.

The zeros of ${x}^{6} + {x}^{3} + 1$ are the Complex $9$th roots that are not $3$rd roots, i.e.

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

with $k = \pm 1 , \pm 2 , \pm 4$

The factors with Real coefficients, composed of Complex conjugate pairs are:

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

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

for $k = 1 , 2 , 4$

Hence:

${x}^{8} + {x}^{7} + {x}^{6} + {x}^{5} + {x}^{4} + {x}^{3} + {x}^{2} + x + 1$

$= \left({x}^{2} + x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{2 \pi}{9}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{4 \pi}{9}\right) x + 1\right) \left({x}^{2} - 2 \cos \left(\frac{8 \pi}{9}\right) x + 1\right)$