Partial fraction expansion - handling multiple roots I am trying to solve the partial fraction expansion of the following: 
$X(z) = \dfrac{3-15z}{9z^2-6z+1}$. 
So I calculated the roots of $9z^2 - 6z + 1$. 
This gives $n_1 = n_2 = \frac{1}{3}$ 
and $9z^2-6z+1 = 9(z-\frac{1}{3})(z-\frac{1}{3})$. 
Now I don't know how to deal with it.
If there were no factor $9$, I would take the approach $X(z) = \frac{A}{z-\frac{1}{3}} + \frac{B}{(z-\frac{1}{3})^2}$. 
And if there were two different roots, $z_1$ and $z_2$, I'd take the approach $X(z) = \frac{A}{z-z_1} + \frac{B}{z-z_2}$. 
But since neither of these two cases is what we have here, I don't know what to do. 

You could absolutely do it your way:  \begin{align}
X(z)=\frac{3-15z}{9z^2-6z+1}&=\frac{3-15z}{9(z-\frac13)^2}\\
&=\frac19\bigg(\frac{3-15z}{(z-\frac13)^2}\bigg)\\
&=\frac19\bigg(\frac{A}{z-\frac13}+\frac{B}{(z-\frac13)^2}\bigg)
\end{align}
Not impossible, but not very nice.
Alternatively:
\begin{align}
X(z)=\frac{3-15z}{9z^2-6z+1}&=\frac{3-15z}{(3z-1)^2}\\
&=\frac{A}{3z-1}+\frac{B}{(3z-1)^2}\\
&=\frac{A(3z-1)+B}{(3z-1)^2}\\
&=\frac{3Az-A+B}{(3z-1)^2}\Rightarrow \left\{\begin{array} 
& 3A &=-15\\
-A+B&=3\end{array} \right. \Rightarrow (A,B)=(-5,-2)\\
&=-\frac{5}{3z-1}-\frac{2}{(3z-1)^2}
\end{align}
