As Fernando Revilla pointed out, we can write this polynomial as [math]\frac{1}{18}\left(12x^5- 3x^3+ 8x^2- 18\right)[/math] so this becomes a question of factoring [math]12x^5- 3x^3+ 8x^2- 18[/math].
By the "rational root theorem", if there is a rational root (so that there is a factor with integer coefficients) then it must be of the form \frac{m}{n} (and the factor of the form (nx- m))where n is an integer factor of the leading coefficient, 12, and m is an integer factor of the constant term, 18. The integer factors of 12 are 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, and -12 and the integer coefficients of 18 are 1, -1, 2, -2, 3, -3, 6, -6, 9, -9, 18, and -18.
Form all the different fractions with those numerators and denominators (there not as many as there might seem- a lot of cancelling occurs) and put them into the polynomial to see If they work.