Chapter 3 - 3.4 - Zeroes of Polynomial Functions - 3.4 Exercises - Page 284: 65

$f(x)=(x+2)(x-2)(x+2i)(x-2i)$ The zeros: $x=-2$ $x=2$ $x=-2i$ $x=2i$

$f(x)=x^4-16$ $f(x)=(x^2)^2-4^2$ $f(x)=(x^2-4)(x^2+4)$ $f(x)=(x^2-2^2)[x^2-(2i)^2]$ $f(x)=(x+2)(x-2)(x+2i)(x-2i)$ The zeros: $x+2=0$ $x=-2$ $x-2=0$ $x=2$ $x+2i=0$ $x=-2i$ $x-2i=0$ $x=2i$