Answer
Possible rational zeros = $\pm1, \pm2, \pm3, \pm6, \pm\dfrac{1}{2}, \pm\dfrac{3}{2}, \pm\dfrac{1}{4}, \pm\dfrac{3}{4}$
Work Step by Step
$f(x)=4x^4-x^3+5x^2-2x-6$
The constant term is -6.
Factors of -6:$ \pm1, \pm2, \pm3, \pm6$
Factors of leading coefficient, 4: $\pm1, \pm2, \pm4$
Possible rational zeros = Factors of -6 $\div$ Factors of 4
Possible rational zeros = $\pm1, \pm2, \pm3, \pm6, \pm\dfrac{1}{2}, \pm\dfrac{3}{2}, \pm\dfrac{1}{4}, \pm\dfrac{3}{4}$
