Question 897477
Something like  {{{a((x^2-something)(x-3))/((x+1)(x-3))}}}


and you want to be able to perform a division (after simplification) that makes a quotient of 2x+1 and have a remainder of denominator of x+1.


The simplification would be to cancel the (x-3) so you have something of {{{a(x^2+- something)/(x+1)}}}; you could ignore the factor "a" in doing the division.   Dividend would be some variable {{{x^2+mx+n}}}, divisor is {{{x+1}}}, and you know that remainder will be some {{{numerator/(x+1)}}}; The quotient as said, must be 2x+1.