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.

$$x^4 + 4x^2 + 16$$
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.
