Chapter 3, Polynomial and Rational Functions - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 329: 25

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

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