Is there a less ad hoc way to find the degree of this extension? I want to find the degree of $\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ over $\mathbb{Q}$. I observe that
$3+2\sqrt{2}=2+2\sqrt{2}+1=(\sqrt{2}+1)^2$ so
$$
\mathbb{Q}(\sqrt{3+2\sqrt{2}})=\mathbb{Q}(\sqrt{2}+1)=\mathbb{Q}(\sqrt{2})
$$
so the degree is 2.
Is there a more mechanical way to show this without noticing the factorization?

We can immediately see that $\sqrt{3+2\sqrt{2}}$ is a root of 
$$(X^2 -3)^2 -8 = X^4 -6X^2 + 1$$
So we can ask, whether this polynomial is irreducible or not.
The polynomial has not roots in $\mathbb Q$, since for a root $\frac rs \in \mathbb Q$ with $(r,s) = 1$, we would need to have $r|1$, $s|1$, but $\pm1$ is not a root.
So let's try to find factors of degree 2:
$$
\begin{align}
X^4 - 6X^2 + 1 &= (X^2 + aX \pm 1)(X^2 + cX \pm 1) \\
&= X^4 + (a+c)X^3 + (ac\pm 2)X^2 \pm (a+c)X + 1
\end{align}
$$
so $a = -c$ and $-6 = ac \pm 2 = -a^2 \pm 2$. This implies that we must choose $a = 2$, $c=-2$ and the minus-sign for $\pm 1$. We have thus arrived at the factorization
$$X^4 - 6X^2 + 1 = (X^2 + 2X - 1)(X^2 - 2X - 1)$$
Both the factors on the right must be irreducible over $\mathbb Q$ (because they can't have a root in $\mathbb Q$) and $\sqrt{3+2\sqrt{2}}$ must be a root of one of them (we need not even figure out which one).
From this it follows that $[\mathbb Q(\sqrt{3+2\sqrt{2}}):\mathbb Q] = 2$.
