Answer
The zeros of the given function are:
$\color{blue}{\left\{1, \dfrac{-1}{2} - \dfrac{\sqrt3i}{2}, \dfrac{-1}{2} + \dfrac{\sqrt3i}{2}\right\}}$
The completely factored form of the given function is:
$\\\color{blue}{P(x) = (x-1)\left(x+\frac{1}{2} + \frac{\sqrt3i}{2}\right)\left(x+\frac{1}{2} - \frac{\sqrt3i}{2}\right)}$
Work Step by Step
RECALL:
(1) $a^3-b^3=(a-b)(a^2+ab+b^2)$
(2) The quadratic formula: $x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$
The given polynomial function can be written as $P(x) = x^3-1^3$.
Factor the difference of two cubes using the formula in (1) above with $a=x$ and $b=1$ to obtain:
$P(x) = (x-1)(x^2+x(1) + 1^2)
\\P(x) = (x-1)(x^2+x+1)$
Equate each factor to zero then solve each equation to obtain:
\begin{array}{ccc}
&x-1=0 &\text{or} &x^2+x+1=0
\\&x=1 &\text{or} &x^2+x+1=0$
\end{array}
Solve the second equation using the formula in (2) above with $a=1, b=1$, and $c=1$ to obtain:
$x=\dfrac{-1 \pm \sqrt{1^2-4(1)(1)}}{2(1)}
\\x=\dfrac{-1\pm\sqrt{1-4}}{2}
\\x=\dfrac{-1\pm\sqrt{-3}}{2}
\\x=\dfrac{-1\pm \sqrt{3(-1)}}{2}
\\x=\dfrac{-1\pm \sqrt3 i}{2}
\\x_1=\dfrac{-1}{2} - \dfrac{\sqrt3i}{2}
\\x_2=\dfrac{-1}{2} + \dfrac{\sqrt3i}{2}$
Thus, the zeros of the given function are:
$\color{blue}{\left\{1, \dfrac{-1}{2} - \dfrac{\sqrt3i}{2}, \dfrac{-1}{2} + \dfrac{\sqrt3i}{2}\right\}}$
The completely factored form of the given function is:
$P(x) = \left(x-1\right)\left[x-\left(\frac{-1}{2} - \frac{\sqrt3i}{2}\right)\right]\left[x-\left(\frac{-1}{2} + \frac{\sqrt3i}{2}\right)\right]
\\\color{blue}{P(x) = (x-1)\left(x+\frac{1}{2} + \frac{\sqrt3i}{2}\right)\left(x+\frac{1}{2} - \frac{\sqrt3i}{2}\right)}$
