Question: Out of idle curiosity, while teaching Calculus II, I started to wonder about this:

How do you factor the polynomial
$$
f\_6(x)=x^6+x^5+x^4+x^3+x^2+x+1
$$
into quadratic and linear factors over $\mathbb R$?

I know how to do it for $f\_5$ (defined in the obvious way), $f\_4$, and lower.

Answer: 
$$f\_6(x)=\frac{x^7-1}{x-1}=\prod\_{k=1}^{3}\left(x^2-2x\cos\frac{2k\pi}{7}+1\right).$$
The polynomial is $\dfrac{x^7-1}{x-1}$, so its roots are
$$
\cos\frac{2k\pi}{7}+i\sin\frac{2k\pi}{7}\qquad (k=1,2,3,4,5,6)
$$
Determine the conjugate pairs and you're done.