Using the theorem we look at the coefficient of the highest power term, and we can see it's -4, which has factors 1*(-4), 4*(-1), 2*(-2), -2*2. Then we look at the constant term, 6: 1*6, -1*(-6), 2*3, -2*(-3). The first set of factors we call the q set and the second set the p set. The rational roots, if any, will be given by p/q, where p and q are members of their respective sets. The set of p/q values is: 1/4, 1/2, 3/4, 1, 3/2, 2, 3, 6, in positive or negative form. We can now try out all possible roots by substitution (the x,y values are shown in pairs from smallest to largest values of x):
-6 -5058; -3 -285; -2 -42; -3/2 -4.5; -1 7; -3/4 7.922; -1/2 7.5; -1/4 6.672;
1/4 5.672; 1/2 5.5; 3/4 4.922; 1 3; 3/2 -10.5; 2 -50; 3 -297; 6 -5082
Although none of these produce zeroes, we can see by the sign changes where the irrational roots lie: between -3/2 and -1; between 1 and 3/2. (Actual values are -1.3789 and 1.1829 approx.)