Answer
zeros: $-2 $ and $1$
$P(x)=(x-1)(x+2)^{2}$
Work Step by Step
$P(x)=x^{3}+3x^{2}-4$
(1 sign variation, we expect 1 positive real zero)
$a_{0}=-14 \qquad $p: $\pm 1,\pm 2,\pm 4$
$a_{n}=1,\qquad $q: $\pm 1, $
Possible rational zeros $\displaystyle \frac{p}{q}:\quad \pm 1,\pm 2,\pm 4$
Test $x=1 $ by synthetic division (with 0 as the missing coefficient) :
$\left[\begin{array}{lllllll}
\underline{1|} & 1 & 3 & 0 & -4 & & \\
& & 1 & 4 & 4 & & \\
& -- & -- & -- & -- & & \\
& 1 & 4 & 4 & \underline{|0} & \Rightarrow & P(1)=0,
\end{array}\right.$
$1$ is a zero, $(x-1) $ is a factor of $P(x).$
The bottom row gives coefficients of the quotient.
$P(x)=x^{3}+3x^{2}-4=(x-1)(x^{2}+4x+4)$
We can factor the second parentheses:
recognize a square of a sum of x and 2
$P(x)=(x-1)(x+2)^{2}$
So, the zeros are $-2 $ and $1$
