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 $(x^2 - 4)(x^2 +4)$?
$\endgroup$ 34 Answers
$\begingroup$That is, since $(x^2+4)(x+2)(x-2)$ is the simplest form of the equation $x^4-16$, rather than $(x^2-4)(x^2+4)$.
$\endgroup$ $\begingroup$Notice that $x^4-16=(x^2)^2-4^2\\ =(x^2-4)(x^2+4)\\ =(x+2)(x-2)(x^2+4)\\ =(x-2)(x+2)(x+2\color{red}i)(x-2\color{red}i)$
$\endgroup$ $\begingroup$They are both factorizations of $x^4-16,$ but $(x^2+4)(x^2-4)$ is a less complete factorization.
$\endgroup$ $\begingroup$Actually you could factor all the way to complex roots:
$$x^4-16 = (x^2-4)(x^2+4) = (x-2)(x+2)(x-2i)(x+2i).$$
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