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.

Yes, there is a "purely mechanical" approach. Given algebraic numbers $\alpha$ and $\beta$, and monic polynomials  $p_1(x)$ and $p_2(x)$ with rational coefficients, of which $\alpha$ and $\beta$ are roots, respectively, we can produce monic polynomials $p_+(x)$ and $p_\times(x)$ with rational coefficients, of which $\alpha+\beta$ and $\alpha\beta$ are roots, respectively. Moreover, if $\alpha$ and $\beta$ are algebraic integers (that is, we can take $p_1,p_2$ to have integer coefficients), then $p_+,p_\times$ have integer coefficients, so they witness that $\alpha+\beta$ and $\alpha\beta$ are algebraic integers as well. The argument is classical, but I follow below the presentation in 

MR1083765 (91i:11001). Niven, Ivan; Zuckerman, Herbert S.; Montgomery, Hugh L. An introduction to the theory of numbers. Fifth edition. John Wiley & Sons, Inc., New York, 1991. xiv+529 pp. ISBN: 0-471-62546-9.

The construction is based on the following lemma:

Lemma. Given $n\ge0$, and a complex number $\xi$, suppose that the complex numbers $\theta_1,\dots,\theta_n$ are not all zero, and satisfy the equations
  $$ \xi\theta_j=a_{j,1}\theta_1+\dots+a_{j,n}\theta_n $$
  for $j=1,2,\dots,n$. If the $n^2$ numbers $a_{j,k}$ are rational, then $\xi$ is algebraic. If they are integers, then $\xi$ is an algebraic integer. 

One proves this by noticing that if $A$ is the matrix of the $a_{j,k}$ and $x$ is the vector of the $\theta_j$, then $Ax=\xi x$, so $\det(A-\xi I)=0$, and this is a monic polynomial with rational coefficients if the $a_{j,k}$ are rational, and integer coefficients if they are integers. In fact, we did better than stated in the lemma, since we obtained a witnessing polynomial rather than simply knowing the numbers are algebraic.
Using the lemma, one proceeds as follows: Suppose that $p_1$, the polynomial for $\alpha$, has degree $m$, and $p_2$, the polynomial for $\beta$, has degree $s$. Consider the numbers $n=ms$ numbers $\alpha^a\beta^b$ with $0\le a\le m-1$ and $0\le b\le s-1$, and call them $\theta_1,\dots,\theta_n$. Note that each $\alpha\theta_j$ is a linear combination of the $\theta_k$, using rational coefficients, and similarly for $\beta\theta_j$. To see this, note that either $\alpha\theta_j$ is another $\theta_i$, or else $\theta_j=\alpha^{m-1}\beta^b$ for some $b$, but then $$\alpha\theta_i=\alpha^m\beta^b=(\alpha^m-0)\beta^b=(\alpha^m-p(\alpha))\beta^b,$$ which is a combination of the $\alpha^i \beta^b$ for $0\le i<m$. The same argument applies to $\beta\theta_j$.
But then it follows that the lemma applies with both $\xi=\alpha+\beta$ and $\xi=\alpha\beta$. And this gives the result. In the case where $\alpha=\sqrt2$ and $\beta=\sqrt3$, this procedure is precisely what MJD sketched in his answer, and results in a polynomial of degree $4$ for $\sqrt2+\sqrt3$. The one thing that is not guaranteed is that in all cases the polynomial we obtain this way is minimal (that is, irreducible over the rationals) if $p_1$ and $p_2$ are minimal. It is in many cases that one finds in practice, though. See this and this MO questions for some details on when this is the case.
