Answer
$x \in \{-3, -1, 2\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f(x)$ with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$.
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$f(x)=x^{3}+2x^2-5x-6$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm3, \pm6$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm3, \pm6$
b. Try for $x=2:$
$\begin{array}{lllll}
\underline{2}| & 1 & 2 & -5 & -6\\
& & 2 & 8 & 6\\
& -- & -- & -- & --\\
& 1 & 4 & 3 & |\underline{0}
\end{array}$
$2$ is a zero,
$f(x)=(x-2)(x^{2} +4x+3)$
c. Factorize the trinomial...find two numbers whose sum is $4$ and whose product is $3$...
$x^2+4x+3=x^2+x+3x+3$...
$x(x+1)+3(x+1)$,
$(x+1)(x+3)=(x+1)(x+3)$
$f(x)=(x-2)(x+3)(x+1)$
The zeros of f satisfy $f(x)=0$
$(x-2)(x+1)(x+3)=0$
$x \in \{-3, -1, 2\}$
