Answer
zeros: $-2, -1,$ and $7$
$P(x)=(x+1)(x+2)(x-7)$
Work Step by Step
$P(x)=x^{3}-4x^{2}-19x-14$
$a_{0}=-14 \qquad $p: $\pm 1,\pm 2,\pm 7,\pm 14$
$a_{n}=1,\qquad $q: $\pm 1, $
Possible rational zeros $\displaystyle \frac{p}{q}:\quad \pm 1,\pm 2,\pm 7,\pm 14$
Test $x=-1 $ by synthetic division:
$\left[\begin{array}{lllllll}
\underline{-1|} & 1 & -4 & -19 & -14 & & \\
& & -1 & 5 & 14 & & \\
& -- & -- & -- & -- & & \\
& 1 & -5 & -14 & \underline{|0} & \Rightarrow & P(-1)=0,
\end{array}\right.$
$1$ is a zero, $(x+1) $ is a factor of $P(x).$
The bottom row gives coefficients of the quotient.
$P(x)=x^{3}-4x^{2}-19x-14=(x+1)(x^{2}-5x-14)$
We can factor the second parentheses:
find factors of $-14$ that add to $-5$ ... ($2$ and $-7$)
$P(x)=(x+1)(x+2)(x-7)$
So, the zeros are $-2, -1,$ and $7$
