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

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 $B=-1$.
