<s>[INST] How to obtain this factorization of $x^4+4$? $x^4 + 4 = (x^2 + 2x +2)(x^2 - 2x +2)$
I am curious how would one obtain this factorization?
Clearly, once the factorization is known it is routine to verify it, however the hard part is how to find the factorization in the first place?
Thanks!

I observed that $(A+B)(A-B)=A^2-B^2$ can be applicable here with $A=x^2+2$, $B=2x$.
Is there any other trick?
 [/INST] This is pretty straight forward if you employ complex numbers and this formula: $$(a+b)(a-b)=a^2-b^2$$
Here we go:
\begin{align} 
x^4 + 4 & = (x^2)^2 - (-4) \\
& = (x^2)^2 - (2i)^2 \\
& = (x^2 + 2i) \ (x^2 - 2i) \\
& = (x+ i\sqrt{2i}) \ (x- i\sqrt{2i}) \ (x+ \sqrt{2i}) \ (x- \sqrt{2i})
\end{align}
We know that $\sqrt{i}=e^{i{\pi\over4}} = {1+i\over\sqrt{2}}$
Thus, 
\begin{align}
x^4 + 4 & = (x+ i\sqrt{2i}) \ (x- i\sqrt{2i}) \ (x+ \sqrt{2i}) \ (x- \sqrt{2i}) \\
& = (x+ 1-i) \ (x- 1+i) \ (x+ 1+i) \ (x- 1-i)\\
&= (x \pm 1 \pm i)
\end{align}
You can also get these roots by using the polar form of complex number, as mentioned by Ross Millikan, or you can also find the roots of the quadratic equations, as written by azc.
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