I'm not sure to be honest. Are you referring to the subtraction on line 189? In any event, I've computed some test cases to see how this might work with floating point numbers.

```julia
(3.4344 x^4 + 3.222)/(1.23 x^2 + 3.2233 x^1 + 4.3234)

# should equal (according to wolframalpha)

(2.7922 x^2 - 7.31714 x + 9.36062) R(1.46284 x - 37.2477)
```

Julia outputs:
```julia
2.7921951219512198x^2 + -7.31714027364664x^1 + 9.360619393334398) R(1.4628397685491201x^1 + -37.247701885141936)
```