Answer
$P(x)=(x-1)(x+2)^{2}$
Zeros: $-2,\ 1\ \ \ $
Work Step by Step
$P(x)=x^{3}+3x^{2}-4$
$P(-x)=-x^{3}+3x^{2}-4$
Decscart's rule of signs:
P(x) has $1$ sign changes $\Rightarrow 1$ positive zeros.
P(-x) has $2$ sign change $\Rightarrow$ 2 or 0 negative zeros.
Rational Zeros Theorem:
candidates for p:$\quad \pm 1,\pm 2,\pm 4$
candidates for q:$\quad \pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 4$
Testing with synthetic division,
$\left.\begin{array}{l}
1|\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 &3& 0&-4\\\hline
&1& 4 &4\\\hline
1& 4& 4&|\ \ 0\end{array}$
Remainder th:
$x=1$ is a zero $\Rightarrow (x-1)$ is a factor (Factor th.).
$P(x)=(x-1)(x^{2}+4x+4)$
For the trinomial, we recognize a perfect square.
$P(x)=(x-1)(x+2)^{2}$
Zeros: $-2,\ 1\ \ \ $
