My question has to do with changing improper rational functions to the sum of a polynomial and proper rational function. If it matters, the context of this is inverse Z-transforms.

My textbook says to do this by doing polynomial long division with the polynomials written in **reverse** order. But I have assignments where the solutions do the long division in the original order, and if I tried doing it in reverse order, I get an answer with different signs that I can't easily factor.

Textbook example (reverse order):

$\displaystyle X(z)= \frac{1 + 3z^{-1} + \frac{11}{6} z^{-2} + \frac{1}{3} z^{-3}}{1 + \frac{5}{6} z^{-1} + \frac{1}{6} z^{-2}}$

$\displaystyle X(z) = 1 + 2 z^{-1} + \frac{\frac{1}{6} z^{-1}}{1 + \frac{5}{6} z^{-1} + \frac{1}{6} z^{-2}}$

Assignment example (normal order):

$\displaystyle H(z) = \frac{1 + z^{-1} + 6 z^{-2}}{1 - z^{-1} - 6 z^{-2}}$

$\displaystyle H(z) = -1 + \frac{2}{1 - z^{-1} - 6 z^{-2}} = -1 + \frac{2}{(1+2z^{-1})(1-3z^{-1})}$

Now, if I do the assignment example using the reverse order, I get:

$\displaystyle H(z) = \frac{1 + z^{-1} + 6 z^{-2}}{1 - z^{-1} - 6 z^{-2}}$

$\displaystyle H(z) = -1 + \frac{2}{1 + z^{-1} + 6 z^{-2}}$

...which I can't factor out the denominator in the nice way it was done in the normal order example.

So what's the deal here? Does the order I do the division not matter, and I should just choose the one that gives me the easier answer to work with? Or is one right and the other wrong?