Answer
$\pm1, \pm3, \pm5, \pm 15,\pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{15}{2},\pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{5}{4}, \pm \frac{15}{4},\pm \frac{1}{8}, \pm \frac{3}{8}, \pm \frac{5}{8}, \pm \frac{15}{8}$
Work Step by Step
Given: $f(x)=8x^4+4x^3-10x+15$
Factors of the constant term: $\pm 1, \pm 3, \pm 5, \pm 15$
Factors of the leading coefficient: $\pm 1, \pm 2, \pm 4, \pm 8$
Possible rational zeros: $\pm \frac{1}{1}, \pm \frac{3}{1}, \pm \frac{5}{1}, \pm \frac{15}{1},\pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{15}{2},\pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{5}{4}, \pm \frac{15}{4},\pm \frac{1}{8}, \pm \frac{3}{8}, \pm \frac{5}{8}, \pm \frac{15}{8}$
Simplified list of possible zeros: $\pm1, \pm3, \pm5, \pm 15,\pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{15}{2},\pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{5}{4}, \pm \frac{15}{4},\pm \frac{1}{8}, \pm \frac{3}{8}, \pm \frac{5}{8}, \pm \frac{15}{8}$
