Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

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 $$ t^{2n}+t^n+1=\prod_{k=0}^{n-1}\Bigl(t^2-2\cos\bigl(\tfrac{2\pi}{3n}+\tfrac{2k\pi}{n}\bigr)t+1\Bigr). $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.