#### Answer

(a) $\left\{0, \dfrac{-1+\sqrt3 i}{2}, \dfrac{-1-\sqrt3 i}{2} \right\}$.
(b) $P(x) = x(x^2+x+1)$

#### Work Step by Step

$\bf{(a) \text{ Zeros}}$
Factor the polynomial completely to obtain:
$P(x) = x(x^2+x+1)$
Equate each factor to zero then solve each equation to obtain:
\begin{array}{ccccc}
&x=0 &\text{or} &x^2+x+1=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 $a=1, b=1, c=1$.
Substitute these values into the quadratic formula above to obtain:
$x=\dfrac{-1\pm \sqrt{1^2-4(1)(1)}}{2(1)}
\\x=\dfrac{-1\pm \sqrt{1-4}}{2}
\\x=\dfrac{-1\pm \sqrt{-3}}{2}
\\x=\dfrac{-1\pm \sqrt{(-1)(3)}}{2}
\\x=\dfrac{-1\pm i\cdot \sqrt3}{2}
\\x=\dfrac{-1\pm \sqrt3 i}{2}$
Thus, the zeros of the function are: $\left\{0, \dfrac{-1+\sqrt3 i}{2}, \dfrac{-1-\sqrt3 i}{2} \right\}$.
$\bf{(b) \text{ Completely Factored Form}}$
From part (a) above, the completely factored form of $P(x)$ is:
$P(x) = x(x^2+x+1)$