Answer
a. The zeros of the function are: $\{1,-2,\frac{-1}{2}+\frac{\sqrt 3}{2}i, \frac{-1}{2}-\frac{\sqrt 3}{2}i, 1 + \sqrt 3i,1 - \sqrt 3i\}$
b. $P(x)=(x-1)(x^{2}+x+1)(x+2)(x^{2}-2x+4)$
Work Step by Step
(a) Zeros
Factor the polynomial completely to obtain:
$P(x)=x^{6}-7x^{3}-8=x^{6}-x^{3}+8x^{3}-8=(x^{6}-x^{3})+(8x^{3}-8)=x^{3}(x^{3}-1)+8(x^{3}-1)=(x^{3}-1)(x^{3}+8)=(x-1)(x^{2}+x+1)(x+2)(x^{2}-2x+4)$
$x-1=0 \rightarrow x=1$
$x+2=0 \rightarrow x=-2$
$x^{2}+x+1=0 \rightarrow x=\frac{-1}{2}\pm\frac{\sqrt 3}{2}i$
$x^{2}-2x+4=0 \rightarrow x=1\pm \sqrt 3i$
Thus, the zeros of the function are: $\{1,-2,\frac{-1}{2}+\frac{\sqrt 3}{2}i, \frac{-1}{2}-\frac{\sqrt 3}{2}i, 1 + \sqrt 3i,1 - \sqrt 3i\}$
(b) Completely Factored Form
From part (a) above, the completely factored form of P(x) is:
$P(x)=(x-1)(x^{2}+x+1)(x+2)(x^{2}-2x+4)$
