Answer
$P(x)=(x-1)(x+2)(x+10)$
Zeros: $ \ -10,\ -2, \ 1 \ $
Work Step by Step
$P(x)=x^{3}+11x^{2}+8x-20$
$P(-x)=-x^{3}+11x^{2}-8x-20$
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:$\quad \pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$
candidates for q:$\quad \pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$
Testing with synthetic division,
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 & 11 & 8 &-20\\\hline
&1 & 12 & 20\\\hline
1& 12 & 20&|\ \ 0\end{array}$
$P(x)=(x-1)(x^{2}+12x+20)$
For the trinomial, find two factors of $+20$ with sum $12$
... ( they are $+2$ and $+10)$
$P(x)=(x-1)(x+2)(x+10)$
Zeros: $ \ -10,\ -2, \ 1 \ $
