Answer
Possible rational zeros = $\pm1, \pm2, \pm3, \pm6, \pm\dfrac{1}{3}, \pm\dfrac{2}{3}$
Work Step by Step
$f(x)=3x^4-11x^3-x^2+19x+6$
The constant term is 6.
Factors of 6:$ \pm1, \pm2, \pm3, \pm6$
Factors of leading coefficient, 3: $\pm1, \pm3$
Possible rational zeros = Factors of 6 $\div$ Factors of 3
Possible rational zeros = $\pm1, \pm2, \pm3, \pm6, \pm\dfrac{1}{3}, \pm\dfrac{2}{3}$
