Let S be the sum of all values of r such that the coefficient of the x^2 term in the expansion of
(x−3)(rx+2)^3 is −6 . If S=a/b, where a and b are coprime positive integers, what is the value of a+b
(x−3)(rx+2)^3
= r^3x^4 - 3r^2(r-2)x^3 - 6r(3r-2)x^2 - 4(9r-2)x - 24
So, we want
-6r(3r-2) = -6
3r^2 - 2r - 1 = 0
(3r+1)(r-1) = 0
r = 1 or -1/3
1 - 1/3 = 2/3
2+3=5