polynomial long division quick question about the algorithm

Question

I was making use of polynomial long division in inverse Z transform and I got stuck in a brainfart in one stage of the polynomial long division.

I posted the original question into digital signal processing stack exchange, but nobody answered it so I thought about sharing the link to math stack exchange.

So here is the link to the my earlier post containing the question and my concerns

https://dsp.stackexchange.com/questions/53426/inverse-z-transform-confused-about-polynomial-long-division-lti-and-causal

if you don't want to look at the link then I can put the same photo here and show the direct question also

I don't know what I should do here, I have two terms (17/2−5z). Which one do you divide by (2z^2) and why is that?

Answer 1

You've gone too far. After the first step, when you have $5z-4$, you should stop because the exponent in $z^2$ is too high for it to go into $5z$. You simply have a remainder of $5z-4$, thus:

$$\frac{4z^2-5z}{2z^2-5z+2} = 2 + \frac{5z-4}{2z^2-5z+2}$$

Technically, you could keep dividing forever to get an infinite series, as you started to do, but usually the idea is to stop as soon as the degree of the current expression is less than that of the divisor. This is like normal division with numbers, stopping at the decimal point rather than getting the infinite decimal expansion.

Answer 2

The long division algorithm for dividing $a(z)$ by $b(z)$ is meant to find the unique polynomials $q(z)$ and $r(z)$ such that

$$a(x) = q(z)b(z) + r(z)$$ with the degree of $r(z)$ strictly less than the degree of $b(z)$.

In your case after one step you already have this situation for $q(z)=2$ (constant) and $r(z) = 5z-4$ which has degree 1 < degree $b(z)$ which is $2$. So you're done after one step.