It's #(x^3-3x^2+9x-29)/8#, with remainder #87#.

First way to do it is a standard long division algorithm, I guess you know that. In short, you divide #x^4# with #8x# (getting #x^3/8#), then multiply that with #8x+24# (getting #x^4+3x^3#), and subtract that from #x^4-2x# (getting #-3x^3-2x#). Then you do the same thing with that result instead of starting numerator, until you get a single number that cannot be divided by #8x# - and that number is #87#.

Second way is more interesting, and uses Horner's algorithm. It's easier to apply, but harder to see why it works. It works only when denominator is #x-alpha# for some #alpha#, so we'll divide by #x+3# (#alpha# being #-3#), and then divide by #8# separately.

You make a table with #1, 0, 0, -2, 0# (coefficients of numerator) in the first row, and in the second row, drop #1#, multiply by #alpha(=-3)# and add #0# (getting #-3#, then multiply that by #alpha# and add next #0# (getting #9#), multiply that by #alpha# and add #-2# (getting #-29#), and multiply that by #alpha# and add the last #0# (getting #87#). Entries except the last in the second row are the coefficients of quotient (remember that they still have to be divided by 8), and the last entry is the remainder.

The whole process is explained at http://en.wikipedia.org/wiki/Synthetic_division.