Chapter 3, Polynomial and Rational Functions - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 329: 30

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

Factor the polynomial completely to obtain: $P(x)=x^{6}-729$ $P(x)=(x^{3}-27)(x^{3}+27)$ $P(x)=(x-3)(x^{2}+3x+9)(x+3)(x^{2}-3x+9)$ $P(x)=(x-3)(x+3)[x+(\frac{3}{2}+\frac{3\sqrt 3}{2}i)][x+(\frac{3}{2}-\frac{3\sqrt 3}{2}i)][x-(\frac{3}{2}+\frac{3\sqrt 3}{2}i)][x-(\frac{3}{2}-\frac{3\sqrt 3}{2}i)]$ Equate each unique factor to zero then solve each equation to obtain: $x-3=0 \rightarrow x=3$ $x+3=0 \rightarrow x=-3$ $x+(\frac{3}{2}+\frac{3\sqrt 3}{2}i)=0 \rightarrow x=\frac{-3}{2}-\frac{3\sqrt 3}{2}i$ $x+(\frac{3}{2}-\frac{3\sqrt 3}{2}i)=0 \rightarrow x=\frac{-3}{2}+\frac{3\sqrt 3}{2}i$ $x-(\frac{3}{2}+\frac{3\sqrt 3}{2}i)=0 \rightarrow x=\frac{3}{2}+\frac{3\sqrt 3}{2}i$ $x-(\frac{3}{2}-\frac{3\sqrt 3}{2}i)=0 \rightarrow x=\frac{3}{2}-\frac{3\sqrt 3}{2}i$ The zeros of the function are: $\{3,-3,\frac{-3}{2}-\frac{3\sqrt 3}{2}i,\frac{-3}{2}+\frac{3\sqrt 3}{2}i,\frac{3}{2}-\frac{3\sqrt 3}{2}i,\frac{-3}{2}+\frac{3\sqrt 3}{2}i\}$ $x =-3$ with multiplicity 1 $x =3$ with multiplicity 1 $x =\frac{-3}{2}-\frac{3\sqrt 3}{2}i$ with multiplicity 1 $x =\frac{-3}{2}+\frac{3\sqrt 3}{2}i$ with multiplicity 1 $x =\frac{3}{2}+\frac{3\sqrt 3}{2}i$ with multiplicity 1 $x =\frac{3}{2}-\frac{3\sqrt 3}{2}i$ with multiplicity 1