College Algebra, Chapter 4, 4.4, Section 4.4, Problem 20

Determine all rational zeros of the polynomial $P(x) = x^3 - x^2 - 8x + 12$, and write the polynomial in factored form.

The leading coefficient of $P$ is $1$, so all the rational zeros are integers:

They are divisors of the constant term $12$. Thus, the possible candidates are

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

Using Synthetic Division







We find that $1, 3, 4$ and $6$ are not zeros but that $2$ is a zero and that $P$ factors as

$x^3 - x^2 - 8x + 12 = (x - 2)(x^2 + x - 6)$

We now factor $x^2 + x - 6$ using trial and error, the factors are $(x - 2)(x + 3)$, so


$\begin{equation} \begin{aligned} x^3 - x^2 - 8x + 12 =& (x - 2)(x - 2)(x + 3) \end{aligned} \end{equation}$


Therefore, the zeros of $P$ are $2$ and $-3$.