Answer
$P(x)=(x+2)(x+3)(x-5)$
Zeros: $ \ -3,\ -2, \ 5 \ $
Work Step by Step
$P(x)=x^{3}-19x-30$
$P(-x)=-x^{3}+19x-30$
Descart's rule of signs:
P(x) has 1 sign changes $\Rightarrow$ 1 positive zeros.
P(-x) has 3 sign changes $\Rightarrow$ 3 or 1 negative zeros.
Rational Zeros Theorem:
candidates for p:$\quad \pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30$
candidates for q:$\quad \pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 30$
Testing with synthetic division,
$\left.\begin{array}{l}
-2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 & 0 & -19 &-30\\\hline
&-2 & 4 & 30\\\hline
1& -2 & -15&|\ \ 0\end{array}$
$P(x)=(x+2)(x^{2}-2x-15)$
For the trinomial, find two factors of $-15$ with sum $-2$
... ( they are $-5$ and $+3)$
$P(x)=(x+2)(x+3)(x-5)$
Zeros: $ \ -3,\ -2, \ 5 \ $
