Answer
zeros: $-5, 1,$ and $2$
$P(x)=(x-1)(x-2)(x+5)$
Work Step by Step
$P(x)=x^{3}+2x^{2}-13x+10$
$a_{0}=10 \qquad $p: $\pm 1,\pm 2,\pm 5,\pm 10$
$a_{n}=1,\qquad $q: $\pm 1, $
Possible rational zeros $\displaystyle \frac{p}{q}:\quad \pm 1,\pm 2,\pm 5,\pm 10$
Test $x=1 $ by synthetic division:
$\left[\begin{array}{lllllll}
\underline{1|} & 1 & 2 & -13 & 10 & & \\
& & 1 & 3 & -10 & & \\
& -- & -- & -- & -- & & \\
& 1 & 3 & -10 & \underline{|0} & \Rightarrow & P(1)=0,
\end{array}\right.$
$1$ is a zero, $(x-1) $ is a factor of $P(x).$
The bottom row gives coefficients of the quotient.
$P(x)=x^{3}+2x^{2}-13x+10=(x-1)(x^{2}+3x-10)$
We can factor the second parentheses:
find factors of -10 that add to +3 ... (5 and -2)
$P(x)=(x-1)(x-2)(x+5)$
So, the zeros are $-5, 1,$ and $2$
