Answer
Possible rational zeros = $\pm1, \pm2, \pm\dfrac{1}{2}, \pm\dfrac{1}{4}$
Work Step by Step
$f(x)=4x^5-8x^4-x+2$
The constant term is 2.
Factors of 2:$ \pm1, \pm2$
Factors of leading coefficient, 4: $\pm1, \pm2, \pm4$
Possible rational zeros = Factors of 2 $\div$ Factors of 4
Possible rational zeros = $\pm1, \pm2, \pm\dfrac{1}{2}, \pm\dfrac{1}{4}$
