Minimal Polynomial of $a +b\sqrt{2}$ as a function of a, b ∈ $\mathbb Q$ 
Determine the minimal polynomial over $\mathbb Q$ of $a +b\sqrt{2}$ as a function of a, b ∈ $\mathbb Q$.

Let $x=a+b\sqrt{2}$ 
If $b=0$ then the minimal polynomial is $x-a$
if not, then $x-a=b\sqrt{2}\iff(x-a)^2-2b^2=x^2-2ax+a^2-2b^2=0$
Is the polynomial further reducible, can I use Eisenstein ?
$p|-2a$ and $p^2\not|a^2-2b^2$
for $p=2, a=4, b=2$ it does not work, or does it ? 

The easiest way is just high-school mathematics, I think: Take $(X-(a+b\sqrt2))(X-(a-b\sqrt2))=X^2-2aX+(a^2-2b^2)$. The discriminant is $(-2a)^2-4\cdot1\cdot(a^2-2b^2)=8b^2$, which is not a square unless $b=0$. In that reducible case, then the two factors I wrote are equal, so you take one of them, $X-a$.
