Answer
$P(x)=(x-2)(x+3)(x-5)$
Zeros: $ \ -3,\ 2, \ 5 \ $
Work Step by Step
$P(x)=x^{3}-4x^{2}-11x+30$
$P(-x)=-x^{3}-4x^{2}+11x+30$
Descart's rule of signs:
P(x) has 2 sign changes $\Rightarrow$ 2 or 0 positive zeros.
P(-x) has 1 sign changes $\Rightarrow$ 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,
$\left.\begin{array}{l}
2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 & -4 & -11 &30\\\hline
& 2 & 4 & -30\\\hline
1& 2 & -15&|\ \ 0\end{array}$
$P(x)=(x-2)(x^{2}+2x-15)$
For the trinomial, find two factors of $-15$ with sum $2$
... ( they are $+3$ and $-5)$
$P(x)=(x-2)(x+3)(x-5)$
Zeros: $ \ -3,\ 2, \ 5 \ $
