College Algebra, Chapter 4, 4.5, Section 4.5, Problem 8

a.) Find all zeros of $P(x) = x^3 + x^2 + x$ of $P$, real and complex

b.) Factor $P$ completely.



a.) We first factor $P$ as follows.


$\begin{equation} \begin{aligned} P(x) =& x^3 + x^2 + x && \text{Given} \\ \\ =& x (x^2 + x + 1) && \text{Factor out } x \end{aligned} \end{equation}$


We find the zeros of $P$ by setting each factor equal to :

Setting $x = 0$, we see that $x = 0$ is a zero. More over, setting $x^2 + x + 1 = 0$, by using quadratic formula, we get


$\begin{equation} \begin{aligned} x =& \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ \\ =& \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2 (1)} \\ \\ =& \frac{-1 \pm \sqrt{-3}}{2} \\ \\ =& \frac{-1 \pm \sqrt{3} i}{2} \end{aligned} \end{equation}$


So the zeros of $P$ are $\displaystyle 0, \frac{-1 + \sqrt{3} i}{2}$ and $\displaystyle \frac{-1 - \sqrt{3} i}{2}$.

b.) By complete factorization,


$\begin{equation} \begin{aligned} P(x) =& (x) \left[ x - \left( \frac{-1 + \sqrt{3} i}{2} \right) \right] \left[ x - \left( \frac{-1 - \sqrt{3} i}{2} \right) \right] \\ \\ =& x \left[ x + \left( \frac{1 - \sqrt{3} i}{2} \right) \right] \left[ x + \left( \frac{1 + \sqrt{3} i}{2} \right) \right] \end{aligned} \end{equation}$