Factoring $x^4 - 16$

Question

I was following a calculus tutorial that factored the equation $x^4-16$ into $(x^2 +4) (x+2)(x-2)$.

Why is the factorization of $x^4-16 = (x^2 + 4)(x+2)(x-2)$ rather than $(x^2 - 4)(x^2 +4)$?

Answer 1

That is, since $(x^2+4)(x+2)(x-2)$ is the simplest form of the equation $x^4-16$, rather than $(x^2-4)(x^2+4)$.

Answer 2

Notice that $x^4-16=(x^2)^2-4^2\\ =(x^2-4)(x^2+4)\\ =(x+2)(x-2)(x^2+4)\\ =(x-2)(x+2)(x+2\color{red}i)(x-2\color{red}i)$

Answer 3

They are both factorizations of $x^4-16,$ but $(x^2+4)(x^2-4)$ is a less complete factorization.

Answer 4

Actually you could factor all the way to complex roots:

$$x^4-16 = (x^2-4)(x^2+4) = (x-2)(x+2)(x-2i)(x+2i).$$