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.

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