ab=1⇒a=1 or b=0 or a=−1 if b is even
So we can just solve three different cases.
Case 1: a=1.
x2−11x+29x2−11x+28(x−4)(x−7)x1x2=1=0=0=4=7
Case 2: b=0.
6x2+x−2(3x+2)(2x−1)x3x4=0=0=−23=12
Case 3: a=−1 and b is even.
x2−11x+29x2−11x+30(x−5)(x−6)x5x6=−1=0=0=5=6(inadmissible)
The possible answers are therefore x= -2/3, 1/2, 4, 6, 7