Answer
$P(x)=(x-1)(x+1)(x+3)$
Zeros: $ \ -3,\ -1, \ 1 \ $
Work Step by Step
$P(x)=x^{3}+3x^{2}-x-3$
$P(-x)=-x^{3}+3x^{2}+x-3$
Descart's rule of signs:
P(x) has 1 sign changes $\Rightarrow$ 1 positive zeros.
P(-x) has 2 sign changes $\Rightarrow$ 2 or 0 negative zeros.
Rational Zeros Theorem:
candidates for p:$\quad \pm 1,\pm 3$
candidates for q:$\quad \pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 3$
Testing with synthetic division,
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 & 3 & -1 &-3\\\hline
&1 & 4 & 3\\\hline
1& 4 & 3&|\ \ 0\end{array}$
$P(x)=(x-1)(x^{2}+4x+3)$
For the trinomial, find two factors of $+3$ with sum $4$
... ( they are $+1$ and $+3)$
$P(x)=(x-1)(x+1)(x+3)$
Zeros: $ \ -3,\ -1, \ 1 \ $
