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 $-3$.
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