Chapter 3, Polynomial and Rational Functions - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 329: 24

$P(x)=x[x-(\frac{1}{2}+\frac{\sqrt 3}{2}i)][x+(\frac{1}{2}-\frac{\sqrt 3}{2}i)]$ The zeros of the function are: $\{0,\frac{1}{2}-\frac{\sqrt 3}{2}i,\frac{1}{2}+\frac{\sqrt 3}{2}i\}$ $x =0$ with multiplicity 1 $x =\frac{1}{2}-\frac{\sqrt 3}{2}i$ with multiplicity 1 $x = \frac{1}{2}+\frac{\sqrt 3}{2}i$ with multiplicity 1

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