Consider that the function has a zero, and that any "rearrangement" that expresses the same function must also exhibit this zero. If, in the new expression, x does not appear in the numerator, then the only way for the function to provide the zero, at least as a limit, is for the denominator to go to infinity. But the only way to do that with finite values for x is for a term of the denominator to itself be rational and to contain x in its denominator.
But this is not really different from the clearly unwanted solution where num/denom is rewritten as 1/(num^-1 denom), which is trivial and I think misses the point.
