# find decomposition of polynomial

How to find a decomposition of the following polynomial

$f := t^{2n} + t^n + 1 \in \mathbb{R}[t], n \in \mathbb{N}$

where the decomposition is a product of $n$ normed polynomials of degree 2?

-
To factor over the reals, it sometimes helps to factor of the complex numbers first, and then combine complex conjugate roots. What values can $t^n$ have? – Marc van Leeuwen Jan 10 '12 at 14:57
@MarcvanLeeuwen: What do you mean by factor over the complex numbers? Factor over $(1+0i)t^{2n}$? Well $t^{2n}$ can only have positive values. – meinzlein Jan 10 '12 at 16:20
No I mean factor the whole polynomial $t^{2n} + t^n + 1$ but interpreted as a polynomial with complex coefficients (which happen to be all $0$ or $1$). If $t$ takes a complex value, $t^{2n}$ does not have to be positive, it can be any complex number. But what I meant is: if a complex number $t$ satisfies $t^{2n} + t^n + 1=0$, what precise (complex) value(s) can $t^n$ have? Then you can deduce what values $t$ itself can have, and find all the complex roots of your polynomial, and in fact a factorisation into degree $1$ polynomials. Then combine to get quadratic real polynomials. – Marc van Leeuwen Jan 10 '12 at 16:32
I'm a blockhead ... I'll see if I can sort it out ... – meinzlein Jan 10 '12 at 16:47

Let $P(t)=t^2+t+1=(t-e^{2\pi i/3})(t-e^{-2\pi i/3})$. Then $$t^{2n}+t^n+1=P(t^n).$$ It follows that the roots of $t^{2n}+t^n+1$ are the $n$-th roots of $e^{2\pi i/3}$ and its conjugates (which are the $n$-th roots of $e^{-2\pi i/3}$). The $n$-th roots of $e^{2\pi i/3}$ are $$e^{\bigl(\tfrac{2\pi}{3n}+\tfrac{2k\pi}{n}\bigr)i},\quad 0\le k\le n-1.$$ Putting it all together we get $$t^{2n}+t^n+1=\prod_{k=0}^{n-1}\Bigl(t^2-2\cos\bigl(\tfrac{2\pi}{3n}+\tfrac{2k\pi}{n}\bigr)t+1\Bigr).$$