Answer
$P(x)=(x+1)(x+2)(x-7)$
Zeros: $-2,\ -1, \ 7 \ \ \ $
Work Step by Step
$P(x)=x^{3}-4x^{2}-19x-14$
$P(-x)=-x^{3}-4x^{2}+19x-14$
Decscart's rule of signs:
P(x) has $1$ sign changes $\Rightarrow 1$ positive zeros.
P(-x) has $2$ sign changes $\Rightarrow$ 2 or 0 negative zeros.
Rational Zeros Theorem:
candidates for p:$\quad \pm 1,\pm 2,\pm 7,\pm 14$
candidates for q:$\quad \pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 7,\pm 14$
Testing with synthetic division,
$\left.\begin{array}{l}
-1|\\ \\ \\ \end{array}\right.\begin{array}{rrr}
-1 &-4&-19&-14\\\hline
&-1& 5 &14\\\hline
1& -5&-14&|\ \ 0\end{array}$
Remainder th:
$x=-1$ is a zero $\Rightarrow (x+1)$ is a factor (Factor th.).
$P(x)=(x+1)(x^{2}-5x-14)$
For the trinomial, find two factors of $-14$ with sum $-5$
... ( they are $+2$ and $-7)$
$P(x)=(x+1)(x+2)(x-7)$
Zeros: $-2\ -1, \ 7 \ \ \ $
