Finding the minimal polynomial of $2+\sqrt3$ over $\mathbb{Q}$ Minimal polynomial of $2+\sqrt3$ over $\mathbb{Q}$.
$(2+\sqrt3)^2$= $7+4\sqrt3$
$\implies$ $((2+\sqrt3)^2-7)^2$=$48$
$\implies$ $((2+\sqrt3)^4-14(2+\sqrt3)^2+1=0$
So $2+\sqrt3$ is a root of $x^4-14x^2+1$. Did i do this correct so far? Any tips on factoring this into quadratics? Thanks!
 A: An idea: put $\;x=2+\sqrt3\;$ and now "rationalize" this expression:
$$x-2=\sqrt3\implies x^2-4x+4=3\implies x^2-4x+1=0$$
and thus the above quadratic is the minimal polynomial of $\;2+\sqrt3\;$ , and not the quartic you wrote.
The above method is useful many times to find at least some polynomial vanishing at some number...
A: This strategy can be applied in lots of situations:
\begin{align}
\alpha = 2+\sqrt 3 &\implies \alpha - 2 = \sqrt 3\\
&\implies (\alpha-2)^2 = 3\\
&\implies \alpha^2 - 4\alpha + 1 = 0
\end{align}
so $\alpha$ vanishes on $x^2-4x+1$. It is irreducible since it has no rational roots, so it is the minimal polynomial.
Your $x^4-14x^2+1$ can be factored as $(x^2-4x+1)(x^2+4x+1)$ and it has roots $\pm 2\pm\sqrt 3$. 
To get that factorization, you can first observe that your polynomial has no rational roots (and therefore no linear factors), so it's of the form $(x^2+ax+b)(x^2+cx+d)$. Expanding and comparing coefficients would give you factorization.
A: It can't be the  minimal polynomial since $[\mathbf Q(\sqrt 3):\mathbf Q]=2$, so the minimal polynomial of any element in this extension field has degree $\le 2$.
Here's how to get it: set $x=2+\sqrt3\iff x-2=\sqrt3$, so $x$ satisfies the equation
$$(x-2)^2=3\iff x^2-4x+1=0.$$
You can check this polynomial is irreducible on $\mathbf Q$ (it has no rational root), so it is the minimal polynomial of $2+\sqrt 3$.
