<s>[INST] Factoring quartic equation $$x^4+6x^2+25=0$$
How could I factor it into $(x^2-2x+5)(x^2+2x+5)=0$?
I got the result looking into horrible formulas on wikipedia, but I suppose there's a much easier way. Could you help me?
 [/INST] Notice that
$$
x^4+6x^2+25=(x^4+6x^2+9)+16=(x+3)^2+4^2=(x^2+3+4i)(x^2+3-4i) \quad \forall x\in \mathbb{C}.
$$
Let $u \in \mathbb{C}$ such that $u^2=3+4i$. It is easy to show that $u\in\{2+i,-2-i\}$. We have
\begin{eqnarray}
x^4+6x^2+25&=&(x^2+u^2)(x^2+\bar{u}^2)=(x+iu)(x-iu)(x+i\bar{u})(x-i\bar{u})\\
&=&(x+iu)\overline{(x+iu)}(x-iu)\overline{(x-iu)}\\
&=&|x+iu|^2|x-iu|^2=|x-1+2i|^2|x+1-2i|^2\\
&=&[(x-1)^2+4][(x+1)^2+4]=(x^2-2x+5)(x^2+2x+5).
\end{eqnarray}
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