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.
 A: HINT:
$$(x^2)^2+4^2+4x^2=  (x^2+4)^2-4x^2$$
A: 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: 
$x^6-64 = (x^3+8)(x^3-8) = (x+2)(x^2-2x+4)(x-2)(x^2+2x+4)$
A: In general, if we have $x^4 + (2a - b^2)x^2 + a^2$,  we can factor it as $(x^2 + a)^2 - (bx)^2 = (x^2 - bx + a)(x^2 + bx + a)$. 
On a related note, this factorization is the intuition behind the Sophie-Germain Identity:
$$x^4 + 4y^4 = (x^2 + 2y^2 + 2xy)(x^2 + 2y^2 - 2xy)$$
Indeed, plug in $a = 2y^2, b = 2y$ to get the result.
A: $$x^4 + 4x^2 + 16$$
Let $y = x^2$, the polynomial is then equal to
$$y^2 + 4y + 16$$
Then, use the quadratic formula:
$$y = \frac{-4 \pm \sqrt{4^2 - 4(1)(16)}}{2(1)}$$
$$y = \frac{-4 \pm \sqrt{-48}}{2}$$
$$y = 2 \pm 2i\sqrt{3}$$
Return to $x$
$$x^2 = 2 \pm 2i\sqrt{3}$$
So, we can now convert $x^4 + 4x^2 + 16$ into factors.
$$x^4 + 4x^2 + 16 = \left(x^2 - {2 + 2i\sqrt{3}}\right)\left(x^2 - {2 - 2i\sqrt{3}}\right)$$
Repeat quadratic formula for each factor.
$$x_{left} = \frac{0 \pm \sqrt{0^2 - 4(1)(2 + 2i\sqrt{3})}}{2}$$
$$x_{left} = \pm \frac{\sqrt{-8 - 8i\sqrt{3}}}{2}$$
$$x_{right} = \frac{0 \pm \sqrt{0^2 - 4(1)(2 - 2i\sqrt{3})}}{2}$$
$$x_{right} = \pm \frac{\sqrt{8i\sqrt{3} - 8}}{2}$$
So, we can now convert $x^4 + 4x^2 + 16$ into factors again.
$$x^4 + 4x^2 + 16 \\ = \left(x - \frac{\sqrt{-8 - 8i\sqrt{3}}}{2}\right)\left(x + \frac{\sqrt{-8 - 8i\sqrt{3}}}{2}\right)\\ \left(x - \frac{\sqrt{8i\sqrt{3} - 8}}{2}\right)\left(x + \frac{\sqrt{8i\sqrt{3} - 8}}{2}\right)$$
None of this is too outlandish for a student to know how to do.
Edit: Corrected the mistake in the quadratic formula.
Edit: Fixed the mistake with naively square-rooting the y-roots.
