## Prove that a given polynomial is irreducible over QQ

Show that $p(x) = x^3 + x^2 - 2x - 1$ is irreducible over $\mathbb{Q}$. Use the fact (to be proven later) that $\alpha = 2 \mathsf{cos}(2\pi/7)$ is a root of $p$ to argue that the regular 7-gon is not constructible by straightedge and compass.

Using the rational root theorem, any rational roots of $p(x)$ must be either 1 or -1. Certainly then $p(x)$ has no rational roots. Since $p(x)$ has degree 3, it is irreducible over $\mathbb{Q}$ if and only if it has no roots (in $\mathbb{Q}$), so that $p(x)$ is irreducible. In particular, the roots of $p(x)$ lie in degree 3 extensions of $\mathbb{Q}$.

Suppose now that the regular 7-gon is constructible. The exterior angles of a regular 7-gon have measure $2\pi/7$, so in particular, if the regular 7-gon is constructible, then so is the point $(\mathsf{cos}(2\pi/7), \mathsf{sin}(2\pi/7))$, so that the number $2\mathsf{cos}(2\pi/7)$ is constructible. But we have seen that any constructible element must have degree a power of 2 over $\mathbb{Q}$. Given that $\alpha$ is a root of $p(x)$, this yields a contradiction.