Answer
a) The zeros of the function are: $\{1,-1,\frac{-1}{2}+\frac{\sqrt 3}{2}i, \frac{-1}{2}-\frac{\sqrt 3}{2}i, \frac{1}{2}+\frac{\sqrt 3}{2}i, \frac{1}{2}-\frac{\sqrt 3}{2}i\}$
b)
Work Step by Step
(a) Zeros
Factor the polynomial completely to obtain:
$P(x)=x^{6}-1=(x^{3}-1)(x^{3}+1)=(x-1)(x^{2}+x+1)(x+1)(x^{2}-x+1)$
$x-1=0 \rightarrow x=1$
$x+1=0 \rightarrow x=-1$
$x^{2}+x+1=0 \rightarrow x=\frac{-1}{2}\pm\frac{\sqrt 3}{2}i$
$x^{2}-x+1=0 \rightarrow x=\frac{1}{2}\pm\frac{\sqrt 3}{2}i$
Thus, the zeros of the function are: $\{1,-1,\frac{-1}{2}+\frac{\sqrt 3}{2}i, \frac{-1}{2}-\frac{\sqrt 3}{2}i, \frac{1}{2}+\frac{\sqrt 3}{2}i, \frac{1}{2}-\frac{\sqrt 3}{2}i\}$
(b) Completely Factored Form
From part (a) above, the completely factored form of P(x) is:
$P(x)=(x-1)(x^{2}+x+1)(x+1)(x^{2}-x+1)$
