Use synthetic division:
I will write it up a little verbosely to make the process as clear as possible. Essentially it's like doing long division.
First notice that #3x(2x - 1) = 6x^2-3x#, which will allow us to separate the #x^2# term:
#(6x^2 - x + 6) - 3x(2x - 1)#
#= (6x^2 - x + 6) - (6x^2-3x)#
#= 6x^2 - x + 6 - 6x^2 + 3x#
#=2x + 6#
So #3x# makes a good multiplier, with remainder #(2x+6).
To match the leading term #2x# in the remainder, the next multiplier we want is #1#:
#1xx(2x-1) = 2x-1#
Then
#(2x + 6) - (2x - 1) = 2x + 6 - 2x + 1 = 7#
This gives #7# as our final remainder. If the remainder was zero then we would have a simple factorization.
Add the multipliers that we have found together to get the term #(3x+1)#.
So to summarize where we have got to so far:
#(6x^2 - x + 6) = (3x + 1)(2x - 1) + 7#
Therefore
#(6x^2 - x + 6)/(2x - 1) = (3x + 1) + 7/(2x - 1)#
#= 3x + 1 + 7/(2x - 1)#
