How to find the irreducible polynomial? It is giving me a lot of trouble, and I'm beginning to think it's not possible. 

Find $\operatorname{irr}(2\sqrt{2} + \sqrt{7})$.

I start like this:
$x = 2\sqrt{2} + \sqrt{7}$
I have squared this many times, and I can't ever seem to get the coefficient of the square root to equal a value that is good enough to eliminate the square root. Is there a theorem for things like this?
 A: You have:$$(x-2\sqrt{2})^2 = 7\to x^2-4x\sqrt{2}+1=0\to \left(\dfrac{x^2+1}{4x}\right)^2 = 2\to x^4+2x^2+1 = 32x^2\to x^4-30x^2+1=0$$, and this is the irreducible polynomial you are looking for...
A: The roots of the irreducible polynomial will be $\pm2\sqrt{2}\pm\sqrt{7}$, making the polynomial
$$(x-(2\sqrt{2}+\sqrt{7}))\cdot(x-(2\sqrt{2}-\sqrt{7})) \cdot(x-(-2\sqrt{2}+\sqrt{7}))\cdot (x-(-2\sqrt{2}-\sqrt{7}))$$
$$=x^4-30x^2+1$$
How do we know what the roots will be? Let $f=\operatorname{irr}(2\sqrt{2}+\sqrt{7})\in\mathbb{Q}[x]$. Now consider it as an element of $\mathbb{Q}(\sqrt{2},\sqrt{7})[x]$. It's easy to show that
$$G=\mathrm{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{7})/\mathbb{Q})=\{1,\sigma,\rho,\sigma\rho\}\cong(\mathbb{Z}/2\mathbb{Z})^2,$$
where elements $\sigma$ and $\rho$ defined by
$$\sigma(\sqrt{2})=-\sqrt{2},\;\sigma(\sqrt{7})=\sqrt{7}\quad\quad\rho(\sqrt{2})=\sqrt{2},\;\rho(\sqrt{7})=-\sqrt{7}$$
We can extend automorphisms of a field $K$ to automorphisms of $K[x]$ in the obvious way (acting on coefficients). Because $f\in\mathbb{Q}[x]$, its coefficients are fixed by all elements of $G$, and therefore $f$ itself is fixed by all elements of $G$. But
$$f=(x-(2\sqrt{2}+\sqrt{7}))g$$
for some $g\in\mathbb{Q}(\sqrt{2},\sqrt{7})[x]$. Therefore
$$f=\sigma(f)=\sigma(x-(2\sqrt{2}+\sqrt{7}))\sigma(g)=(x-(-2\sqrt{2}+\sqrt{7}))\sigma(g)$$
demonstrating that $(x-(-2\sqrt{2}+\sqrt{7}))$ is also a factor of $f$, i.e. $-2\sqrt{2}+\sqrt{7}$ is also a root of $f$. Repeating this process gets all four variations of $\pm2\sqrt{2}\pm\sqrt{7}$, and you check at the end that the polynomial you get has rational coefficients.
More generally, if $K$ is a Galois extension of $F$ and some irreducible $f\in F[x]$ has a root in $K$, then in fact $f$ splits in $K$ and $\mathrm{Gal}(K/F)$ permutes the roots of $f$ transitively, so that if $\alpha\in K$ is known to be a root of $f$, then the roots of $f$ will be $\{\psi(\alpha):\psi\in \mathrm{Gal}(K/F)\}$ (possibly with repetitions).
