Answer
$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3} , \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm3$
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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$f(x)=12x^8-x^7+6x^4-x^3+x-3$
a. candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm 3$
$q:\qquad \pm 1, \pm 2,\pm3,\pm4,\pm6,\pm12$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3} , \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm3$
