College Algebra 7th Edition

$P(x)=(x-4)[x+(2+2\sqrt 3i)][x+(2-2\sqrt 3i)]$ The zeros of the function are: $\{4,-2+2\sqrt 3i,-2-2\sqrt 3i\}$ $x =-2+2\sqrt 3i$ with multiplicity 1 $x =-2-2\sqrt 3i$ with multiplicity 1 $x =4$ with multiplicity 1
Factor the polynomial completely to obtain: $P(x)=x^{3}-64$ $P(x)=(x-4)(x^{2}+4x+16)$ $P(x)=(x-4)[x+(2+2\sqrt 3i)][x+(2-2\sqrt 3i)]$ Equate each unique factor to zero then solve each equation to obtain: $x-4=0 \rightarrow x=4$ $x+(2+2\sqrt 3i)=0 \rightarrow x=-2-2\sqrt 3i$ $x+(2-2\sqrt 3i)=0 \rightarrow x=-2+2\sqrt 3i$ The zeros of the function are: $\{4,-2+2\sqrt 3i,-2-2\sqrt 3i\}$ $x =-2+2\sqrt 3i$ with multiplicity 1 $x =-2-2\sqrt 3i$ with multiplicity 1 $x =4$ with multiplicity 1