Computing minimal polynomial An example in chapter 14.2 in Dummit and Foote computes the minimal polynomial for $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$. We consider the field $\mathbb{Q}(\sqrt{2}+\sqrt{3})$, and note that this is the same field as $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Now, the other roots of the minimal polynomial for $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$ are the distinct conjugates of $\sqrt{2}+\sqrt{3}$ under the Galois group. The distincr conjugates are $\pm\sqrt{2}\pm\sqrt{3}$. The minimal polynomial is therefore $[x-(\sqrt{2}+\sqrt{3})][x-(\sqrt{2}-\sqrt{3})][x-(-\sqrt{2}+\sqrt{3})][x-(-\sqrt{2}-\sqrt{3})]$. The next sentence states that this is "quickly computed" to be the polynomial $x^4-10x^2+1$. My question is how is this quickly computed? I'm not sure of a way to do it other than tediously expanding it. Further, how do we know that $x^4-10x^2+1$ is, in fact, irreducible? 

if $x  = \sqrt 2 + \sqrt 3,$ then $x^2 = 5 + 2 \sqrt 6.$ So $(x^2 - 5)^2 = 24,$ and $x^4 - 10 x^2 + 25 = 24,$ and
$$ x^4 - 10 x^2 + 1 = 0. $$
By the Gauss lemma, without any rational roots, the possible rational factorings are
$$ (x^2 + ax +1)(x^2 - ax + 1),  $$
$$ (x^2 + ax -1)(x^2 - ax - 1).  $$
Neither one gives integer $a.$ Either $2 - a^2 = -10$ or $-2 - a^2 = -10$
