"Given that m - 2 is a positive integer that divides 3m^2 - 2m + 11, find the sum of all such values of m."
\(\frac{3m^2-2m+11}{m-2}=3m+4+\frac{19}{m-2}\)
For the last term to be an integer we must have m - 2 = 1, so m = 3; or m - 2 = 19, so m = 21
Hence sum = 3 + 21 = 24