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\begin{document}
\title{Factor $x^{10}+x^{5}+1$}
\author{Amit Yaron}

\maketitle
You may' see a lot of such equations on YouTube. The way to factor
those polynomial is to use the identity 

$x^{3}-1=(x-1)(x^{2}+x+1)$

Let us start by subtracting and adding $x^{7}$:

$x^{10}+x^{5}+1=x^{10}-x^{7}+x^{7}+x^{5}+1$

Now, to continue, let us do the same with $x^{4}$and $x^{2}$:

$x^{10}+x^{5}+1=x^{10}-x^{7}+x^{7}-x^{4}+x^{5}-x^{2}+x^{4}+x^{2}+1=$

$=(x^{3}-1)x^{7}+(x^{3}-1)x^{4}+(x^{3}-1)x^{2}+x^{4}+x^{2}+1=$

$=(x^{3}-1)(x^{7}+x^{4}+x^{2})+(x^{4}+x^{2}+1)\quad(1)$

Now, what is $x^{4}+x^{2}+1$?

Let us add and subtract $x^{2},$then use the formula for difference
of squares:

$x^{4}+x^{2}+1=x^{4}+2x^{2}+1-x^{2}=(x^{2}+1)^{2}-x^{2}=(x^{2}+x+1)(x^{2}-x+1)$

Plug it into (1):

$x^{10}+x^{5}+1=(x-1)(x^{2}+x+1)(x^{7}+x^{4}+x^{2})+(x^{2}+x+1)(x^{2}-x+1)=$

$=(x^{2}+x+1)[(x-1)(x^{7}+x^{4}+x^{2})+x^{2}-x+1]=$

$=(x^{2}+x+1)(x^{8}-x^{7}+x^{5}-x^{4}+x^{3}-x^{2}+x^{2}-x+1)=$

$=(x^{2}+x+1)(x^{8}-x^{7}+x^{5}-x^{4}+x^{3}-x+1)$

You can factor it better if you switch to complex variables. Let us
substitute:

$u=x^{5}$

Then,

$x^{10}+x^{5}+1=u^{2}+u+1=u^{2}+2u\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1=(u+\frac{1}{2})^{2}+\frac{3}{4}=$

$=(u+\frac{1}{2})+(\frac{\sqrt{3}}{2})^{2}=[u-(-\frac{1}{2}+\frac{\sqrt{3}i}{2})][u-(-\frac{1}{2}-\frac{\sqrt{3}i}{2})]=$

$=[u-(\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3}))][u-(\cos(-\frac{2\pi}{3})+i\sin(-\frac{2\pi}{3}))]$

So, the zeros are:

$\cos(\pm\frac{2\pi}{15}+\frac{2n\pi}{5})+i\sin(\pm\frac{2\pi}{15}+\frac{2n\pi}{5})\quad\forall n\in\mathbb{Z}$

Now, the sum of a complex number and its conjugate is real. The same
goes for their product, so:

$[x-(\cos(\frac{2\pi}{15}+\frac{2n\pi}{5})+i\sin(\frac{2\pi}{15}+\frac{2n\pi}{5}))][x-(\cos(-\frac{2\pi}{15}-\frac{2n\pi}{5})+i
\sin(-\frac{2\pi}{15}-\frac{2n\pi}{5}))]=$

$=[x-(\cos(\frac{2\pi}{15}+\frac{2n\pi}{5})+i\sin(\frac{2\pi}{15}+\frac{2n\pi}{5}))][x-(\cos(\frac{2\pi}{15}+\frac{2n\pi}{5})-i
\sin(\frac{2\pi}{15}+\frac{2n\pi}{5}))]=$

$=[x^{2}-2\cos(\frac{(2+6n)\pi}{15})x+1]$

Thus, the complete factorization over $\mathbb{R}$ is:

$x^{10}+x^{5}+1=\prod\limits _{n=0}^{5}[x^{2}-2\cos(\frac{(2+6n)\pi}{15})+1]$

That one was not included in the YouTube video, and is \textbf{\textcolor{red}{ad-free}}!
\end{document}