How do you divide #(x^3+2x^2-11x-12)/(x^2-3x+2)#?

You can just divide the coefficients like this:

The process is similar to long division of numbers.

Note that if there were any 'missing' powers of #x# in the dividend or divisor then we would have to include #0#'s for them.

Write the dividend #1,2,-11,-12# under the bar and the divisor #1,-3,2# to the left.

Choose the first term #color(blue)(1)# of the quotient so that when multiplied by the divisor, the resulting leading term (#1#) matches the leading term (#1#) of the dividend.

Write the product #1,-3,2# of this first term of the quotient and the divisor under the dividend and subtract to give a remainder #5,-13#.

Bring down the next term #-12# from the dividend alongside it to give your running remainder #5,-13,-12#.

Choose the next term #color(blue)(5)# of the quotient so that when multiplied by the divisor, the resulting leading term (#5#) matches the leading term (#5#) of the remainder.

Write the product #5,-15,10# of this second term of the quotient and the divisor under the running remainder and subtract to give a remainder #2,-22#.

There are no more terms to bring down from the dividend, so this is our final remainder.

We find:

#(x^3+2x^2-11x-12)/(x^2-3x+2) = x+5+(2x-22)/(x^2-3x+2)#