Answer
$P(x)=(x-5)(x+3)(x+2)$
Zeros: $-3, -2$ and $5$
Work Step by Step
$P(x)=x^{3}-19x-30,$
(1 sign variation, we expect 1 positive real zero)
$P(-x)=-x^{3}+19x-30$
(2 sign variations, we expect 0 or 2 negative real zeros)
$a_{0}=-30, $p: $\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10, \pm 15,\pm 30$
$a_{n}=1,\qquad $q: $\pm 1, $
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
$\pm 1,\pm 2,\pm 3,\pm 5,\pm,\pm 10, \pm 15,\pm 30$
Test $x= 5 $ by synthetic division:
$\left[\begin{array}{lllllll}
\underline{5|} & 1 & 0 & -19 & -30 & & \\
& & 5 & 25 & 30 & & \\
& -- & -- & -- & -- & & \\
& 1 & 5 & 6 & \underline{|0} & \Rightarrow & P(5)=0,
\end{array}\right.$
$P(x)=(x-5)(x^{2}+5x+6)$
To factor the trinomial, search for two factors of $+6$ whose sum is $+5.$
(These are $+3$ and $+2)$
$P(x)=(x-5)(x+3)(x+2)$
Zeros: $-3, -2$ and $5$
