Unit impulse response $h[n]$ of a discrete time system with multiple characteristic roots 0 I am trying to obtain the unit impulse response of a system in the form:
$$y[n+N]+...+a_{N-1}y[n+1]+a_{N}y[n]=b_0x[n+N]+...+b_{N-1}x[n+1]+b_{N}x[n]$$
$$Q[E]y[n]=P[E]x[n]$$
(where E is the unit advance operator)
Using the following formula:
$$h[n]=\frac{b_N}{a_N}\delta[n]+y_c[n]u[n]$$
Or if $a_N=0$
$$h[n]=A_0\delta[n]+A_1\delta[n-1]+y_c[n]u[n]$$
Where $A_0$ and $A_1$ are determined through iterative calculation of initial values of $h[n]$ and $y_c[n]$ is a combination of the systems characteristic roots that satisfies $Q[E]y_c[n]u[n]=0$
The system in question is:  $y[n]=0.5y[n-1]+x[n]-(0.5)^4x[n-4]$
Upon converting the system to using the operator E notation to obtain the characteristic roots, I obtain 4 roots 3 of which are repeated zeros.
$$(E^4-0.5E^3)y[n]=(E^4-0.5^4)x[n]$$
$$\gamma_{1,2,3} = 0, \gamma_4=0.5$$
Using the relations described above and by simply discarding the repeated zero roots I then obtain the answer:
$$h[n] = (0.5)^nu[n]$$
However, when replacing $x[n]$ and $y[n]$ with $\delta[n]$ and $h[n]$ respectively in the original system difference equation, I obtain:
$$h[n] = (0.5)^n(u[n]-u[n-4])$$
Which is matching with my model answer.
I figured that maybe there is a mistake in my original solution due to the discarding of the repeated zero roots. How should I approach solving systems with zero repeated roots?
If you have a difference equation
$$\sum_{k=0}^Na_ky[n-k]=\sum_{k=0}^Mb_kx[n-k]\tag{1}$$
and if $M&gt;N$, the general expression for the corresponding causal impulse response is
$$h[n]=A_0\delta[n]+A_1\delta[n-1]+\ldots+A_{M-N}\delta[n-(M-N)]+u[n]\sum_{k}c_k\gamma_k^n\tag{2}$$
where $\gamma_k$ are the roots of the characteristic equation, and where I've assumed simple roots. Note that roots $\gamma_k=0$ do not contribute to $(2)$ (even if there are multiple roots equal to zero).
For the given example we have $M-N=3$ and one non-zero root $\gamma=\frac12$. Hence the solution has the form
$$h[n]=A_0\delta[n]+A_1\delta[n-1]+A_2\delta[n-2]+A_3\delta[n-3]+c\left(\frac12\right)^n\tag{3}$$
The constants $A_i$, $i=0,1,\ldots,3$, and $c$ must be determined by iterating through the difference equation with $x[n]=\delta[n]$, $h[n]=y[n]$, and $h[-1]=0$.
If you do that you'll find that $c=0$, which means that the given difference equation describes an FIR system. In terms of transfer functions this means that there is a pole-zero cancellation, leaving only poles at the origin of the complex plane.