Irreducible polynomial of $\sqrt{2}+\sqrt{7}$ on $\Bbb{Q}$.

Question

I would like to find the irreducible polynomial on $\Bbb{Q}$ of $\sqrt{2}+\sqrt{7}$. How can I do that ? First time I see this kind of question, I can find a polynomial $X^2-2$ witch $\sqrt{2}$ is a roots, but I don't understand what I need.

To conclude that the polynomial is irreducible I can use Eisentein criterion but can someone explain how can I find the polynomial ?

Answer 1

Hint : $a=\sqrt{2}+\sqrt{7}$ satisfies $(a-\sqrt{2})^2-7=0$, or $a^2-5=2\sqrt{2}a$, so $(a^2-5)^2=8a^2$