Represent $x^{24}+1$ as a product of irreducible polynomials with real coefficients I am trying to factor the polynomial $f(x)=x^{24}+1$ into a product of irreducibles over $\mathbb{R}$. 
So far I have $x^{24}+1=(x^8+1)(x^{16}-x^8+1)$.
I can show that $g(x)=x^8+1$ is irreducible over $\mathbb{R}$ as follows:
I know that $g(x)$ is irreducible over $\mathbb{R}$ if and only if $g(x+c)$ is irreducible over  $\mathbb{R}$ for any $c \in \mathbb{R}$. Consider
$$g(x+1)=(x+1)^8+1=\sum_{k=0}^8 \binom{8}{k}x^k +1=x^8+\sum_{k=1}^7 \binom{8}{k}x^k + 2.$$
Note that $2$ divides all of the coefficients, except the leading coefficient, but the constant term is not divisible by $2^2=4$. Thus, by Eisenstein's Criterion, $g(x)=x^8+1$ is irreducible over $\mathbb{R}$.
How do I factor $x^{16}-x^8+1$ as a product of irreducibles over $\mathbb{R}$?
Thank you for your help!
 A: The complex roots of this polynomial are the 24th roots of $-1$, which are:
$$\cos(\theta_k) \pm i \sin(\theta_k); \qquad \theta_k = \frac{\pi(2k + 1)}{24}; \qquad 0 \leq k \leq 11$$
So the polynomial $x^{24} + 1$ factors over $\mathbb{R}$ into quadratics of the form $(x - \alpha)(x - \overline{\alpha})$, where $\alpha$ and $\overline{\alpha}$ are complex conjugate roots. So we get:
$$x^{24} + 1 = \prod_{k = 0}^{11}(x - (\cos(\theta_k) + i\sin(\theta_k))(x - (\cos(\theta_k) - i\sin(\theta_k)) = \prod_{k = 0}^{11}(x^2 -2\cos(\theta_k)x + 1)$$
This might be enough for your purposes, but you can also express $\cos(\theta_k)$ as an algebraic expression using various trig identities like the half-angle identity.
A: The factorization you are looking for is
$$x^{24}+1=\prod_{k=1}^{12}\left(x^2-2\cos\left({2k-1\over24}\pi\right)x+1\right)$$
All that's going on is that the equation $x^{24}=-1$ implies $x=e^{\pi in/24}$ for an odd value of $n$.  Pairing off roots $e^{\pi in/24}$ and $e^{-\pi in/24}$ with $n=2k-1$ gives the quadratic
$$(x-e^{\pi in/24})(x-e^{-\pi in/24})=x^2-2\left(e^{\pi in/24}+e^{\pi in/24}\over2 \right)x+1=x^2-2\cos\left({\pi n\over24}\right)x+1$$
and letting $k$ run from $1$ to $12$ picks up all the distinct roots.
Ah, I see Alex Zorn posted essentially the same answer while I was writing this up.
