Answer
$x\in\{2-i, 2+i, 3\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f$ 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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$f(x)=x^{3}-7x^{2}+17x-15$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 3, \pm5, \pm15$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 3,\pm5, \pm15$
b. Try for $x=3:$
$\begin{array}{lllll}
\underline{3}| & 1 & -7 & 17 & -15\\
& & 3 & -12 & 15\\
& -- & -- & -- & --\\
& 1 & -4 & 5 & |\underline{0}
\end{array}$
$3$ is a zero,
$f(x)=(x-3)(x^{2} -4x+5)$
c. Solve for the trinomial using the quadratic formula for the quadratic equation of $ax^2+bx+x$, $x=\frac{-b\pm \sqrt {b^2-4ac}}{2a}$,
in this case, $x^2-4x+5$, $x=\frac{4\pm \sqrt {(-4)^2-4\times 1\times 5}}{2\times 1}=\frac{4\pm 2i}{2}=2 \pm i$
$x\in\{2-i, 2+i, 3\}$
