Chapter 3, Polynomial and Rational Functions - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 329: 23

$P(x)=x(x+2i)(x-2i)$ The zeros of the function are: $\{0,2i,-2i\}$ $x =0$ with multiplicity 1 $x =2i$ with multiplicity 1 $x = -2i$ with multiplicity 1

Factor the polynomial completely to obtain: $P(x)=x^{3}+4x$ $P(x)=x(x^{2}+4)=x(x+2i)(x-2i)$ Equate each unique factor to zero then solve each equation to obtain: $x=0$ $x+2i=0 \rightarrow x=-2i$ $x-2i=0 \rightarrow x=2i$ The zeros of the function are: $\{0,2i,-2i\}$ $x =0$ with multiplicity 1 $x =2i$ with multiplicity 1 $x = -2i$ with multiplicity 1