College Algebra, Chapter 4, 4.5, Section 4.5, Problem 52

Determine all the zeros of the polynomial $P(x) = 2x^3 - 8x^2 + 9x - 9$.
Based from the theorem, the possible rational are the factors of $9$ divided by the factors of the leading coefficient 2 which are $\displaystyle \pm \frac{1}{1}, \pm \frac{3}{1}, \pm \frac{9}{1}, \pm \frac{1}{2}, \pm \frac{3}{2} \text{ and } \pm \frac{9}{2}$. Thus, by using synthetic division and trial and error,


Hence,

$\begin{equation} \begin{aligned} x &= \frac{-(-2) \pm \sqrt{(-2)^2-4(2)(3)}}{2(2)}\\ \\ &= \frac{2\pm\sqrt{-20}}{4}\\ \\ &= \frac{2\pm 2\sqrt{-5}}{4}\\ \\ &= \frac{1 \pm \sqrt{5}i}{2} \end{aligned} \end{equation}$


To find the complex roots, we use quadratic formula.

$\begin{equation} \begin{aligned} x &= \frac{-2 \pm \sqrt{2^2 - 4(1)(3)}}{2(1)}\\ \\ &= \frac{-2 \pm \sqrt{-8}}{2} = \frac{-2\pm2\sqrt{-2}}{2} = -1\pm\sqrt{2}i \end{aligned} \end{equation}$

Thus, the zeros of $P$ are $\displaystyle 3, \frac{1 + \sqrt{5}i}{2} \text{ and } \frac{1 - \sqrt{5}i}{2}$