Answer
$\pm 1, \pm 2, \pm 3, \pm 4,\pm 6,\pm 12,\frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{1}{6}$
Work Step by Step
Given: $f(x)=6x^3-3x^2+15$
Factors of the constant term: $\pm 1, \pm 2, \pm 3, \pm 4,\pm 6,\pm 12$
Factors of the leading coefficient: $\pm 1, \pm 2, \pm 3, \pm 6$
Possible rational zeros: $\pm \frac{1}{1}, \pm \frac{2}{1}, \pm \frac{3}{1}, \pm \frac{4}{1}, \pm \frac{6}{1},\pm \frac{12}{1},\pm \frac{1}{2}, \pm \frac{2}{2}, \pm \frac{3}{2}, \pm \frac{4}{2}, \pm \frac{6}{2},\pm \frac{12}{2},\pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{3}{3}, \pm \frac{4}{3}, \pm \frac{6}{3},\pm \frac{12}{3},\pm \frac{1}{6}, \pm \frac{2}{6}, \pm \frac{3}{6}, \pm \frac{4}{6}, \pm \frac{6}{6},\pm \frac{12}{6}$
Simplified list of possible zeros: $\pm 1, \pm 2, \pm 3, \pm 4,\pm 6,\pm 12,\frac{1}{2},\pm \frac{3}{2},\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{1}{6}$
