<s>[INST] Factoring $x^4 + 4x^2 + 16$ I was putting together some factoring exercises for my students, and came across one that I am unsure of how to factor.
I factored $x^6 - 64$ as a difference of squares, and then tried it as a difference of cubes, but was left with $(x^2 - 4)(x^4 + 4x^2 + 16)$ is there a general method for factoring $x^4 + 4x^2 + 16$?It factors into two irreducible quadratic trinomials, which is where I think the problem is stemming from.
Thanks in advance.
 [/INST] Here is another method that might generalize to more situations. 
Using DeMoivre's Theorem, the roots of $x^6-64 = 0$ are $x = 2e^{ik\pi/3}$ for $k = 0,1,2,3,4,5$. 
So, the polynomial is the product of the following complex linear factors: 
$(x-2)(x+2)(x-2e^{i\pi/3})(x-2e^{i5\pi/3})(x-2e^{i2\pi/3})(x-2e^{i4\pi/3})$
Now pair up factors that are complex conjugates, and simplify to get: 
$(x-2)(x+2)(x^2-(4\cos\frac{\pi}{3})x+4)(x^2-(4\cos\frac{2\pi}{3})x+4)$
$= (x-2)(x+2)(x^2-2x+4)(x^2+2x+4)$
Alternatively, you can first use the difference of squares factorization, and then use the sum of cubes and difference of cubes factorizations: 
$x^6-64 = (x^3+8)(x^3-8) = (x+2)(x^2-2x+4)(x-2)(x^2+2x+4)$
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