College Algebra, Chapter 4, 4.4, Section 4.4, Problem 74

Determine the integers that are upper and lower bounds for the real zeros of the polynomial $P(x) = 2x^3 - 3x^2 - 8x + 12$
The possible rational zeros of $P$ are $\displaystyle \pm \frac{1}{2}, \pm 1, \pm \frac{3}{2}, \pm 2, \pm, 3, \pm 4, \pm 6, \pm 12$. By testing these numbers,

We find that $2$ is a root of the polynomial and using long division, we have



So,

$\begin{equation} \begin{aligned} 2x^3 - 3x^2 - 8x + 12 &= (x - 2)(2x^2 + x - 6) && \text{Factor $2x^2 + x - 6$ using trial and error}\\ \\ 2x^3 - 3x^2 - 8x + 12 &= (x - 2)(2x - 3)(x + 2) \end{aligned} \end{equation}$

Therefore, the real zeros are $\displaystyle 2, \frac{3}{2} \text{ and } -2 \text{ and } -2$ is the lower bound and $2$ is the upper bound for the real zeros of $P$.