Answer
$\pm1, \pm2, \pm3,\pm 6, \pm 7,\pm 14,\pm 21,\pm 42, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3}$
Work Step by Step
Given: $f(x)=3x^4+5x^3-3x+42$
Factors of the constant term: $\pm 1, \pm 2, \pm 3, \pm 6, \pm 7,\pm 14,\pm 21,\pm 42$
Factors of the leading coefficient: $\pm 1, \pm 3$
Possible rational zeros: $\pm \frac{1}{1}, \pm \frac{2}{1}, \pm \frac{3}{1}, \pm \frac{6}{1}, \pm \frac{7}{1}, \pm \frac{14}{1}, \pm \frac{21}{1}, \pm \frac{42}{1}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{3}{3}, \pm \frac{6}{3}, \pm \frac{7}{3}, \pm \frac{14}{3}, \pm \frac{21}{3}, \pm \frac{42}{3},$
Simplified list of possible zeros: $\pm1, \pm2, \pm3,\pm 6, \pm 7,\pm 14,\pm 21,\pm 42, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{7}{3}, \pm \frac{14}{3}$
