The numerator $z^{2}-2 z+1$ can be factored as $(z-1)^{2}$ and the denominator $z^{2}-1$ can be factored as $(z-1)(z+1)$. The polynomial $4 z^{2}-z-3$ can be factored as $(4z+3)(z-1)$.
So, the given expression becomes:
$$\frac{(z-1)^{2}}{(z-1)(z+1)}
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