Answer
$x \in \{-3, 1, 5\}$ all with a multiplicity of $1$
Work Step by Step
See the Rational Zeros 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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$f(x)=x^{3}-3x^2-13x+15$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 3, \pm5, \pm15$
$q:\qquad \pm 1, $
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm3,\pm5,\pm15$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| & 1 & -3 & -13 & 15\\
& & 1 & -2 & -15\\
& -- & -- & -- & --\\
& 1 & -2 & -15 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-2)(x^{2} -2x-15)$
c. Factorizing the trinomial $(x^2-2x-15)$
(find factors of $-15(1)=-15$ whose sum is $-2$):
($-5$ and $+3$).
thus, $x^2-2x-15=x^2+3x-5x-15=x(x+3)-5(x+3)=(x-5)(x+3)$
$f(x)=(x-1)(x-5)(x+3)$,
thus, the zeros are:
$x \in \{-3, 1, 5\}$ all with a multiplicity of $1$
