Constructing a degree 4 rational polynomial satisfying $f(\sqrt{2}+\sqrt{3}) = 0$ Goal: Find $f \in \mathbb{Q}[x]$ such that $f(\sqrt{2}+\sqrt{3}) = 0$.
A direct approach is to look at the following
$$
\begin{align}
(\sqrt{2}+\sqrt{3})^2 &= 5+2\sqrt{6} \\
(\sqrt{2}+\sqrt{3})^4 &= (5+2\sqrt{6})^2 = 49+20\sqrt{6} \\
\end{align}
$$
Putting those together gives
$$
-1 + 10(\sqrt{2}+\sqrt{3})^2 - (\sqrt{2}+\sqrt{3})^4 = 0,
$$
so $f(x) = -1 + 10x^2 - x^4$ satisfies $f(\sqrt{2}+\sqrt{3}) = 0$.
Is there a more mechanical approach? Perhaps not entirely mechanical, but something more abstract.

There is a mechnical procedure, as follows.
Any polynomial function of $r = \sqrt 2 + \sqrt 3$ must have the form $a + b\sqrt 2 + c\sqrt 3 + d\sqrt 6$ for rational $a,b,c,d$.  Consider the set of numbers of that form as a vector space $V$ over the rationals.  It has dimension 4.
Now calculate $r^0, r^1, r^2, r^3, r^4$.  These are five elements of the vector space $V$, and since $V$ has dimension only 4, they cannot be linearly independent. Therefore there must exist rationals $a_0,\ldots, a_4$ such that $a_4r^4 + a_3r^3 + a_2r^2 + a_1r^1 + a_0r^0 = 0$. These can be found by well-known mechanical methods for changing the basis of a vector space.  Then our polynomial is $a_4x^4 + a_3x^3 + a_2x^2 + a_1x^1 + a_0$.
(There are a couple of fine points I skipped here: $a_4$ might be zero; $r^3$ might not be independent of $r^0, r^1, $ and $r^2$. None of this is hard to deal with.) 

Here is an example.
Calculate powers of $r = \sqrt2 + \sqrt3$, and tabulate them:
$$\begin{array}{crrrr}
%   & 1 & \sqrt2 & \sqrt3 &\sqrt 6\\
%\hline
r^0 = & 1 &&&\\
r^1 = & & \sqrt2  & + \sqrt3 & \\
r^2 = & 5 &  && + 2\sqrt6\\ 
r^3 = & &11\sqrt2  &+  9\sqrt3 \\
r^4 = & 49 &&& + 20\sqrt 6
\end{array}$$
Now we want to find rational $a,b,c,d$ such that $r^4 = ar^3 + br^2 + cr^1 + dr^0$.  Such rationals must exist. (Unless $r^0\ldots r^3$ are not independent, in which case we are looking for a polynomial of lower degree, and we can use the same method with even less effort.)  The relations in the table above impose relations on $a,b,c,d$ that we can read off from the table, one relation for each column:
$$
\begin{array}{rrrrl}
& 5b & & + d  &=49\\
11a&& + c &&= 0\\
9a&&+c&&=0\\
&2b&&& = 20
\end{array}
$$
We can solve the equations mechanically (they are particularly simple in this case; you can just read off the answer) and find that $a=0, b=10, c=0, d=-1$.  So we have calculated, entirely mechanically,  that $r^4 = 10r^2-1$, which means that $r$ is a zero of the polynomial $$x^4-10x^2+1.$$
(I wrote this up in detail on my blog a few years back, and just happened to use $\sqrt 2 + \sqrt 3$ as an example.)
