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 ? 
 A: If the polynomial were reducible, it would be reducible into a product of two linear factors, but if $b\ne0,$ then neither of these would have a rational constant term.
A: 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, $X-a$.
A: We need a polynomial $p(x)$ with a root $r=a+b\sqrt{2}$
Now $\frac{r-a}{b}=\sqrt{2}$ and we have $\large(\frac{r-a}{b}\large)^2=2$
Thus replacing $r$ with $x$ we get an irreducible polynomial over $\mathbb{Q}$ satisfied by $r$
$f(x)=\large(\frac{x-a}{b}\large)^2-2$
The minimal polynomial would be 
$g(x)=b^2f(x)=(x-a)^2-2b^2$
As an exercise, you can try finding the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$ which is not so easy to find otherwise (at least it gave me a hard time when I was new to Field Theory)
