How did Euler realize $x^4-4x^3+2x^2+4x+4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$? How did Euler find this factorization?
$$\small x^4 −  4x^3 + 2x^2 + 4x + 4=(x^2-(2+\alpha)x+1+\sqrt{7}+\alpha)(x^2-(2-\alpha)x+1+\sqrt{7}-\alpha)$$
where $\alpha = \sqrt{4+2\sqrt{7}}$
I know that he had some super powers, like he was sent to us from a super intelligent alien universe just to humiliate our intelligence, but how the hell did he do that three centuries ago? :|

I like this rather old question. Here is a yet another possible way Euler could have taken:
Note that $\displaystyle x^4+ax^2+b$ can be factorized easily if $\displaystyle a^2-4b\geq 0$. If, however, $\displaystyle a^2-4b\leq 0$, then 
\begin{align}
x^4+ax^2+b&=(x^2+\sqrt{b})^2-(x\sqrt{2\sqrt{b}-a})^2\\
&=(x^2+\sqrt{b}-x\sqrt{2\sqrt{b}-a})(x^2+\sqrt{b}+x\sqrt{2\sqrt{b}-a}).
\end{align}
Now in $P(x)=x^4-4x^3+2x^2+4x+4$, use $x=y+1$, and proceed:
\begin{align}
P(y+1)&=y^4-4y^2+7\\
&=(y^2+\sqrt{7}-y\sqrt{2\sqrt{7}+4})(y^2+\sqrt{7}+y\sqrt{2\sqrt{7}+4}).
\end{align}
substituting $y=x-1$ we arrive at Euler's result.
