Answer
$a.\displaystyle \quad\pm 1,\pm\frac{1}{5}$ .
$b.\displaystyle \quad-1,\frac{1}{5},$ and $1$
Work Step by Step
Rational Zeros Theorem$:$
$ ... $every rational zero of $P(x)$ is of the form $\displaystyle \frac{p}{q}$
where $p$ and $q$ are integers and
$p$ is a factor of the constant coefficient $a_{0}$
$q$ is a factor of the leading coefficient $a_{n}$
---
$a.$
$P(x)=5x^{3}-x^{2}-5x+1$
candidates for p: $\pm 1,$
candidates for q: $\pm 1,\pm 5$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad $$\displaystyle \pm 1,\pm\frac{1}{5}$ .
$b.$
From the graph, the actual zeroes are $-1,\displaystyle \frac{1}{5},$ and $1.$
