Answer
$P(x)=(x-1)(x+10)(x+2)$
Zeros: $-10, -2$ and $1$
Work Step by Step
$P(x)=x^{3}+11x^{2}+8x-20,$
(1 sign variation, we expect 1 positive real zero)
$P(-x)=-x^{3}+11x^{2}-8x-20,$
(2 sign variations, we expect 0 or 2 negative real zeros)
$a_{0}=-20, $ p: $\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$
$a_{n}=1,\qquad $q: $\pm 1, $
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
$\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20.$
Test $x= 1 $ by synthetic division:
$\left[\begin{array}{lllllll}
\underline{1|} & 1 & 11 & 8 & -20 & & \\
& & 1 & 12 & 20 & & \\
& -- & -- & -- & -- & & \\
& 1 & 12 & 20 & \underline{|0} & \Rightarrow & P(1)=0,
\end{array}\right.$
$P(x)=(x-1)(x^{2}+12x+20)$
To factor the trinomial, search for two factors of $+20$ whose sum is $+12.$
(These are $+10$ and $+2)$
$P(x)=(x-1)(x+10)(x+2)$
Zeros: $-10, -2$ and $1$
