Answer
$P(x)=(x+1)^{2}(x-2)$
Zeros: $-1,\ 2\ \ \ $
Work Step by Step
$P(x)=x^{3}-3x-2$
$P(-x)=-x^{3}+3x-2$
Decscart's rule of signs:
P(x) has $1$ sign changes $\Rightarrow 1$ positive zeros.
P(-x) has $2$ sign changes $\Rightarrow$ 2 or 0 negative zeros.
Rational Zeros Theorem:
candidates for p:$\quad \pm 1,\pm 2$
candidates for q:$\quad \pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2 $
Testing with synthetic division,
$\left.\begin{array}{l}
-1\ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 &0& -3&-2\\\hline
&-1& 1 &2\\\hline
1& -1& -2&|\ \ 0\end{array}$
Remainder th:
$x=-1$ is a zero $\Rightarrow (x+1)$ is a factor (Factor th.).
$P(x)=(x+1)(x^{2}-x-2)$
For the trinomial, find two factors of $-2$ with sum $-1$
... ( they are $-2$ and $+1)$
$P(x)=(x+1)(x+1)(x-2)=(x+1)^{2}(x-2)$
Zeros: $-1,\ 2\ \ \ $
