Chapter 3 - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 293: 23

$P(x)=x(x+2i)(x-2i)$ zeros: $0, -2i, 2i$ (each with multiplicity 1)

Factoring x out of both terms, $P(x)=x(x^{2}+4)$ Zeros of $x^{2}+4$: $x^{2}=-4$ $x=\pm 2i,$ $x^{2}+4=(x+2i)(x-2i)$ so, $P(x)=x(x+2i)(x-2i)$ zeros: $0, -2i, 2i$ (each with multiplicity 1)