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?
This reduces to checking if the radical $\:\sqrt{3+2\sqrt{2}}\:$ denests. While there are general algorithms, simple cases like this can be tackled by employing an easy formula that I discovered as a teenager.
Simple Denesting Rule $\rm\quad \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $
E.g. $\ 3 + 2\sqrt{2}\ $ has norm $\ (3+2\sqrt{2})\:(3-2\sqrt{2})\ =\ 9 - 4\cdot 2\ =\ 1$
So $\rm\:\color{blue}{subtracting\ out}\ \sqrt{norm}\ = \pm1\ $ yields $\rm\ 3 + 2\sqrt{2}\: -\: \pm1\ =\ 2 + 2\sqrt{2}\ \ or\ \ 4 + 2\sqrt{2} $
$\rm\qquad \sqrt{trace(2+2\sqrt{2})}\ =\ \sqrt{4}\ =\ 2,\quad\ \ so\ \quad\ \color{brown}{dividing\ it\ out}\ \ \ (2+2\sqrt{2})/2\ =\ 1+\sqrt{2}$
$\rm\qquad \sqrt{trace(4+2\sqrt{2})}\ =\ \sqrt{8}\ =\ 2\sqrt{2}\quad so\quad \color{brown}{dividing\ it\ out}\ \ \ (4+2\sqrt{2})/(2\sqrt{2})\: =\: \sqrt{2}+1$
Indeed $\rm\ (1 + \sqrt{2})^2\ =\ 3 + 2\sqrt{2}\:.\ $ It is an easy exercise to check that the formula is correct.
With experience from a few worked examples, one can swiftly mentally denest such radicals.
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 $[\mathbb Q(\sqrt{3+2\sqrt{2}}):\mathbb Q] = 2$.
Here's how I would do it:
Observe first that there is a tower of fields $\mathbb{Q}\subset \mathbb{Q}(\sqrt2)\subset \mathbb{Q}\left(\sqrt{3+2\sqrt2}\right)$.
Now, the first extension has degree 2. To determine the degree of the extension you're asking, by transitivity of degrees it suffices to compute the degree of the second extension.
This extension has degree $\leq 2$, since $\left(\sqrt{3+2\sqrt2}\right)^2=3+2\sqrt2\in \mathbb{Q}(\sqrt2)$.
Let us try to prove (or disprove) it is of degree 1, i.e. let us try to find $a,b\in \mathbb{Q}$ such that $\sqrt{3+2\sqrt2}=a+b\sqrt2$.
Squaring both sides, we get $3+2\sqrt2=a^2+2b\sqrt2+2b^2$.
This forces $b=1$ and then $a=1$, so actually $\sqrt{3+2\sqrt2}=1+\sqrt2$.
So $\left\lvert\mathbb{Q}\left(\sqrt{3+2\sqrt2}\right):\mathbb{Q}(\sqrt2)\right\rvert=1$, therefore $\left\lvert \mathbb{Q}\left(\sqrt{3+2\sqrt2}\right):\mathbb{Q}\right\rvert=2$.
