First, a fraction is zero if and only if its numerator is 0. So you are really asking for zeroes of the cubic polynomial
3x3−6x2+x−2. . . . . . By the "rational roots theorem" the only possible rational number solutions are
±2,
±31, and
±32. Right off, we see that
3(23)−6(22)+2−2=24−24+2−2=0. So x= 2 is a solution! That means (x- 2) is a factor:
3x3−6x2+x−2=(x−2)(3x2+1). That quadratic term has no real zeroes but two complex zeroes.