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?
 A: You can try as follows (based on the $(a+b)^2=a^2+2ab+b^2)$):
\begin{align*}
x^4+4&=(x^2)^2+2^2\\
&=\big(x^2+2\big)^2-4x^2\\
&=\big(x^2+2\big)^2-(2x)^2\\
&=(x^2+2+2x)(x^2+2-2x)\\
&=(x^2+2x+2)(x^2-2x+2).
\end{align*}
A: People found a lot of things by trial and error.  If you know complex numbers, it is natural to say $x^4=-4, x=\sqrt[4]{-4}$ and use the polar form to find the four roots $x=\pm 1 \pm i$.  Combine them in conjugate pairs to clear the imaginaries and you have the factorization.
A: This is called as the Sophie Germain Identity.  You can see wikipedia link.
A: 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.
A: Well, it's clear $x^4 = -4$ has 4 complex roots so if it factors at all it (which it might not) it must factor as $(x^2 + bx + c)(x^2 + dx + e) $ [because $(x+a)(x^3 + ...) $ requires $a $ to be complex]
So I need $ec=4$; $eb + cd=0$; $e +  bd + c =0$.  Symmetry compels me to first try $e = c= 2$ (I'll try e=1; c=4 if this fails but the rational root test implies these are the only options).  Then $b=-d $ and $bd=-4$ so (wolog) $b=2$ $d=-2$.
