Answer
Possible rational zeros = $\pm1, \pm3, \pm5, \pm15, \pm\dfrac{1}{2}, \pm\dfrac{3}{2}, \pm\dfrac{5}{2}, \pm\dfrac{15}{2}$
Work Step by Step
$f(x)=2x^4+3x^3-11x^2-9x+15$
The constant term is 15.
Factors of 15:$ \pm1, \pm3, \pm5, \pm15$
Factors of leading coefficient, 2: $\pm1, \pm2$
Possible rational zeros = Factors of 15 $\div$ Factors of 2
Possible rational zeros = $\pm1, \pm3, \pm5, \pm15, \pm\dfrac{1}{2}, \pm\dfrac{3}{2}, \pm\dfrac{5}{2}, \pm\dfrac{15}{2}$
