Answer
$\pm1, \pm2, \pm7,\pm 14, \pm \frac{1}{2}, \pm \frac{7}{2},\pm \frac{1}{4},\pm \frac{7}{4}$
Work Step by Step
Given: $f(x)=4x^5+3x^3-2x-14$
Factors of the constant term: $\pm 1, \pm 2, \pm 7, \pm 14$
Factors of the leading coefficient: $\pm 1, \pm 2,\pm 4$
Possible rational zeros: $\pm \frac{1}{1}, \pm \frac{2}{1}, \pm \frac{7}{1}, \pm \frac{14}{1}, \pm \frac{1}{2}, \pm \frac{2}{2}, \pm \frac{7}{2}, \pm \frac{14}{2},\pm \frac{1}{7}, \pm \frac{2}{7}, \pm \frac{7}{7}, \pm \frac{14}{7} $
Simplified list of possible zeros: $\pm1, \pm2, \pm7,\pm 14, \pm \frac{1}{2}, \pm \frac{7}{2},\pm \frac{1}{4},\pm \frac{7}{4}$
