How would I go about factorising this into irreducibles in $\mathbb{Q}$

Question

How could I factorise $$x^{32} - 5x^{16} + 4$$ I know I could possibly just test integers due to the factor theorem, however that could be difficult due to the degree of the expression, how else could I do it?

Thanks in advance.

Answer 1

Note that $x^{32} - 5x^{16} + 4 = (x^{16} - 4)(x^{16} - 1)$. From that point on, use the irreducibility of cyclotomic polynomials over $\mathbb{Q}$ (or the equivalent).

Answer 2

First of all, it factorizes to $(x^{16}-1)(x^{16}-4)$. The first factor further decomposes to $(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)$ and the second to $(x^8-2)(x^8+2)$.

It is clear that $x^2+1$ is irreducible over rationals. If $x^4+1$ would be reducible, it could only be reducible to a product of two quadratic expressions (otherwise there would be a rational root of $-1$). If you use the decomposition over reals $(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$, you see that these factors are not rational, so $x^4+1$ is irreducible.

A bit more work requires to show that $x^8\pm 2$ and $x^8+1$ are irreducible too: use the decomposition into complex linear factor and check that products of those never result in rational polynomials.