How to find the degree of the extension $[\mathbb{Q}(\sqrt[4]{3+2\sqrt{5}}):\mathbb{Q}]$?

Here's a nice trick how to show that $f(x)=x^8-6x^4-11$ is irreducible over $\mathbb{Q}$.

By Gauss' lemma it is irreducible over $\mathbb{Q}$ iff it is irreducible over $\mathbb{Z}$. Let $f=gh$ be a product of two polynomials with integer coefficients. Looking at the constant term of $f$ which is $-11$, you see that the constant terms of $g$ and $h$ have to be $\pm 1$ and $\pm 11$. Since the constant term is the product of all roots of the polynomial up to a sign, $g$ or $h$, and as a result $f$, has a root $\alpha$ with $\lvert \alpha \rvert \leq 1$. But then $\lvert \alpha^8 - 6 \alpha^4 -11 \rvert \geq 11-6-1 > 0$ and so $\alpha$ is not a zero of $f$, contradiction.

Hence $f$ is irreducible.