#### Answer

(a) $\left\{0, 1-i, 1+i\right\}$.
(b) $P(x) = x(x^2-2x+2)$

#### Work Step by Step

$\bf{(a) \text{ Zeros}}$
Factor the polynomial completely to obtain:
$P(x) = x(x^2-2x+2$)
Equate each factor to zero then solve each equation to obtain:
\begin{array}{ccccc}
&x=0 &\text{or} &x^2-2x+2=0\end{array}
The second equation can be solved using the quadratic formula $x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$.
The trinomial has $a=1, b=-2, c=2$.
Substitute these values into the quadratic formula above to obtain:
$x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(1)(2)}}{2(1)}
\\x=\dfrac{2\pm \sqrt{4-8}}{2}
\\x=\dfrac{2\pm \sqrt{-4}}{2}
\\x=\dfrac{2\pm \sqrt{4(-1)}}{2}
\\x=\dfrac{2\pm 2\sqrt{-1}}{2}
\\x=\dfrac{2\pm 2i}{2}
\\x=1\pm i$
Thus, the zeros of the function are: $\left\{0, 1-i, 1+i\right\}$.
$\bf{(b) \text{ Completely Factored Form}}$
From part (a) above, the completely factored form of $P(x)$ is:
$P(x) = x(x^2-2x+2)$