Answer
a. $\frac{p}{q}:\qquad \pm 1, \pm13$
b. $1$ is a zero of the function
c. $x\displaystyle \in\{\ 2 \pm 3i, 1\}$
Work Step by Step
See The Rational Zero Theorem:
If $\displaystyle \frac{p}{q}$ 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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$f(x)=x^{3}-5x^2+17x-13$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm13$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm13$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| & 1 & -5 & 17 & -13\\
& & 1 & -4 & 13\\
& -- & -- & -- & --\\
& 1 & -4 & 13 & |\underline{0}
\end{array}$
${1}$ is a zero,
$f(x)=(x-1)(x^{2} -4x+13)$
c. Using quadratic formula, $x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$.
we can solve for the quadratic $ax^2+bx+c$,
In our case of, $x^{2} -4x+13$, $a=1, b=-4, c=13$
$x=\frac{4 \pm \sqrt {(-4)^2-4\times 1 \times 13}}{2\times 1}$,
$x=\frac{4 \pm 6i}{2}$,
$x={2 \pm 3i}$
The zeros of f satisfy $f(x)=0$
$x\displaystyle \in\{\ 2 \pm 3i, 1\}$
