Cover-up rule for transfert function:$\frac{z^2+2z-1}{z^2(z-1)}$

Question

$$\frac{Y(z)}{z}=\frac{z^2+2z-1}{z^2(z-1)}=\frac{A}{z-1}+\frac{B}{z}+\frac{C}{z^2}$$

I find $A=2$ but what are the values for $B$ and $C$ ?

From the book Introduction to Digital Signal Processing written by Bob Meddins

Regards

Answer 1

You have obtained $$ \frac{z^2+2z-1}{z^2(z-1)}=\frac{2}{z-1}+\frac{B}{z}+\frac{C}{z^2} \tag1 $$ then multiplying out by $z^2$ one gets $$ \frac{z^2+2z-1}{(z-1)}=\frac{2z^2}{z-1}+z\: B+C $$ letting $z:=0$ gives $C=1$ then one has $$ \frac{z^2+2z-1}{z^2(z-1)}=\frac{2}{z-1}+\frac{B}{z}+\frac{1}{z^2} \tag2 $$letting $z:=-1$ gives
$$ 1=\frac{2}{-2}+\frac{B}{-1}+\frac{1}{1} \tag2 $$ and $B=-1$.