[tex]f(x)=\boxed{1}x^5-3x^2+3x+\boxed{14}[/tex]
The potential rational zeros are divisors of 14:
[tex]\pm1;\ \pm2;\ \pm7;\ \pm14[/tex]
and the quotient p/q where p is a divisor of 1, and q is a divisor of 14:
[tex]\pm\dfrac{1}{2};\ \pm\dfrac{1}{7};\ \pm\dfrac{1}{14}[/tex]
