Answer
a. $\displaystyle \pm 1, \pm\frac{1}{5}$
b. $-1, \displaystyle \frac{1}{5}, 1$
Work Step by Step
If $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$
has integer coefficients,
then all the rational zeros of $P$ have the form $x=\displaystyle \pm\frac{p}{q}$
where $p$ is a divisor of the constant term $a_{0}$ and
$q$ is a divisor of the leading coefficient $a_{n}$.
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a.
$a_{0}=1, \qquad $p: $\pm 1$
$a_{n}=5,\qquad $q: $\pm 1, \pm 5$
Possible $\displaystyle \frac{p}{q}:\quad \pm 1, \pm\frac{1}{5}.$
b.
From the graph, actual zeros (x-intercepts):
$-1, \displaystyle \frac{1}{5}, 1.$
