## College Algebra (6th Edition)

Possible rational zeros = $\pm1, \pm2, \pm3, \pm6, \pm\dfrac{1}{2}, \pm\dfrac{3}{2}, \pm\dfrac{1}{4}, \pm\dfrac{3}{4}$
$f(x)=4x^4-x^3+5x^2-2x-6$ 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 = $\pm1, \pm2, \pm3, \pm6, \pm\dfrac{1}{2}, \pm\dfrac{3}{2}, \pm\dfrac{1}{4}, \pm\dfrac{3}{4}$