Factors of the expression $x^{10}+x^5+1$.

Question

Question: Find the factors of the expression $x^{10}+x^5+1$.

My work :

$x^5=t$ so we have : $t^2+t+1$ now we know $x^3-1=(x-1)(x^2+x+1)$ so $t^2+t+1=\dfrac{t^3-1}{t-1}$ now what do i do ?

Answer 1

$$x^{10}+x^5+1=x^{10}-x+x^5-x^2+x^2+x+1=$$ $$=x(x^3-1)(x^6+x^3+1)+x^2(x^3-1)+x^2+x+1=$$ $$=(x^2+x+1)((x^2-x)(x^6+x^3+1)+x^3-x^2+1)=$$ $$=(x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1).$$

Answer 2

Note that $x^2+x+1$ divides $f(x)$. Well, how did I see this? This is because if $\omega$ is an imaginary cube root of unity, note that $f(\omega)=0$. And a property of $\omega$ is that $\omega^2+\omega +1=0$.

Can you take it from here?

Answer 3

Hint: Solve $t^2+t+1=0$. You will obtain complex root such that $-0.5+i\frac{\sqrt{3}}{2}$ and $-0.5-i\frac{\sqrt{3}}{2}$. Then you can obtain $5$ roots by solving $x^5=-0.5+i\frac{\sqrt{3}}{2}$. The other $5$ roots can be obtained similarly.

Answer 4

Over reals, it's got to be:

$$x^{10}+x^5+1=(x^2-2\cos \frac {2\pi}{15}x+1)(x^2-2\cos\frac {4\pi}{15}x+1)(x^2-2\cos\frac {8\pi}{15}x+1)(x^2-2\cos\frac {10\pi}{15}x+1)(x^2-2\cos\frac {14\pi}{15}x+1) $$

This is because, if you take $\omega=e^{\frac {2\pi i}{15}} $, we have:

$$\begin {align}x^{10}+x^5+1 & =\frac {x^{15}-1}{x^5-1} \\ & =\frac {(x-1)(x-\omega)(x-\overline{\omega})(x-\omega^2)(x-\overline{\omega^2})(x-\omega^3)(x-\overline{\omega^3})(x-\omega^4)(x-\overline{\omega^4})(x-\omega^5)(x-\overline{\omega^5})(x-\omega^6)(x-\overline{\omega^6})(x-\omega^7)(x-\overline{\omega^7})}{(x-1)(x-\omega^3)(x-\overline{\omega^3})(x-\omega^6)(x-\overline{\omega^6})} \\ & = (x-\omega)(x-\overline{\omega})(x-\omega^2)(x-\overline{\omega^2})(x-\omega^4)(x-\overline{\omega^4})(x-\omega^5)(x-\overline{\omega^5})(x-\omega^7)(x-\overline{\omega^7})\end {align} $$

and note that, for $k=1,2,4,5,7$ we can use: $(x-\omega^k)(x-\overline{\omega^k})=x^2-(\omega^k+\overline {\omega^k})x+\omega^k\overline {\omega^k}=x^2-2\Re(\omega^k)+1=x^2-2\cos\frac {2k\pi}{15}x+1$

Fun fact: a lot of "interesting" trigonometric identities can be obtained by plugging concrete values for $x $ in the above polynomial identity. For example, plugging $x=i $ results in: $$\frac {1}{16}=\cos 24^\circ\cos 48^\circ\cos 96^\circ\cos 168^\circ$$