Answer
zeros: $-1 $ and $2$
$P(x)=(x-2)(x+1)^{2}$
Work Step by Step
$P(x)=x^{3}-3x-2$
(1 sign variation, we expect 1 positive real zero)
$a_{0}=-2 \qquad $p: $\pm 1,\pm 2,$
$a_{n}=1,\qquad $q: $\pm 1, $
Possible rational zeros $\displaystyle \frac{p}{q}:\quad \pm 1,\pm 2$.
Test $x=1 $ by synthetic division (with 0 as the missing coefficient) :
$\left[\begin{array}{lllllll}
\underline{1|} & 1 & 0 & -3 & -2 & & \\
& & 1 & 1 & -2 & & \\
& -- & -- & -- & -- & & \\
& 1 & 1 & -2 & \underline{|-4} & \Rightarrow & P(1)=-4,
\end{array}\right.$
$1$ is not a zero, test $x=2$
$\left[\begin{array}{lllllll}
\underline{2|} & 1 & 0 & -3 & -2 & & \\
& & 2 & 4 & 2 & & \\
& -- & -- & -- & -- & & \\
& 1 & 2 & 1 & \underline{|0} & \Rightarrow & P(2)=0,
\end{array}\right.$
$x=2$ is a zero, $(x-2)$ is a factor of $P(x).$
The bottom row gives coefficients of the quotient.
$P(x)=x^{3}-3x-4=(x-2)(x^{2}+2x+1)$
We can factor the second parentheses:
recognize a square of a sum of x and $1$
$P(x)=(x-2)(x+1)^{2}$
So, the zeros are $-1 $ and $2$