Inverse z-transform. Where is mistake? I've already wrote about that trouble (link here), but I don't understand where I've made a mistake. 

Full description of the task is as follows:

Z-transform of sequence {x(k)} describe by the equation. 
$$
X(z) = \frac{2.5 - 3.15z^{-1} + 1.2z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}} 
$$

Calculate of samples x(-2), x(-1), x(0), x(1) and x(2) double-side of sequence {x(k)}, which correspond to this z-transform.
Note -
Double-side sequence is sequence to consist samples before zero and after zero. 

Step 1

I calculate of the integer part of the expression (just divided the numerator by denominator),

$$
X(z) = \frac{2.5 - 3.15z^{-1} + 1.2z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}}\qquad(1.1)
$$
transform to,
$$
X(z) = 2.5 + \frac{2.6z^{-1} - 1.8z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}}\qquad(1.2)
$$

Step 2

I transformed denominator like $$\frac{(z – p1)(z – p2)}{z^{-2}}\qquad(2.1)$$.
$$
1 - 2.3z^{-1} + 1.2z^{-2} = (1 - 1.5z^{-1})(1 - 0.8z^{-1})\qquad(2.2)
$$

Step 3

I transformed of the expression from:
$$
X(z) = 2.5 + \frac{2.6z^{-1} - 1.8z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}}\qquad(3.1)
$$
to
$$
X(z) = 2.5 + X{_1}(z)\qquad(3.2)
$$
and then $$X{_1}(z)\qquad(3.3)$$ transform to
$$
X{_1}(z) = \frac{A}{1 - 1.5z^{-1}} + \frac{B}{1 - 0.8z^{-1}}\qquad(3.4)
$$

Step 4

Calculate of coefficient A and B

Calculate of A

$$
(1 - 1.5z^{-1})X{_1}(z) = A + \frac{B(1 - 1.5z^{-1})}{1 - 0.8z^{-1}},\qquad(4.1)\qquad if\qquad z=1.5
$$
hence
$$
A = \frac{2.6z^{-1} - 1.8z^{-2}}{1 - 0.8z^{-1}},\qquad(4.2)\qquad if\qquad z = 1.5
$$
so
$$
A = 2
$$

Calculate B

$$
(1 - 0.8z^{-1})X{_1}(z) = \frac{A(1 - 0.8z^{-1})}{1 - 1.5z^{-1}},\qquad(4.3)\qquad if\qquad z=0.8
$$
hence
$$
B = \frac{2.6z^{-1} - 1.8z^{-2}}{1 - 1.5z^{-1}},\qquad(4.4)\qquad if\qquad z = 0.8
$$
so
$$
B = -0.5
$$

Step 5

$$
X{_1}(z) = \frac{2}{1 - 1.5z^{-1}} + \frac{-0.5}{1 - 0.8z^{-1}}\qquad(5.1)
$$
and
$$
X(z) = 2.5 + \frac{2}{1 - 1.5z^{-1}} + \frac{-0.5}{1 - 0.8z^{-1}}\qquad(5.2)
$$
Sequence of {x(k)} has the following representation
$$
x(k) = 2(1.5)^{k}, if k&lt;0;
x(k) = 2.5, if k=0;
x(k) = -0.5(0.8)^{k}, if k&gt;0;
$$
And after that we can calculate Sequence of {x(k)} - x(-2), x( 1), x(0), x(1) и x(2)
 x(-2) = -0.88889;
 x(-1) = -1.33333;
 x(0) = 2.50;
 x(1) = -0.40;
 x(2) = -0.32.

Could you check my solve and say where I made mistake.
Thank you very much!
$$X_1(z) = \frac{2.6z^{-1} - 1.8z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}} =\frac{A}{1 - 1.5z^{-1}} + \frac{B}{1 - 0.8z^{-1}} \tag{1}$$


  Mistake is that $A \neq 2$ and $B \neq -0.5$. 


Well then, what are they ? There aren't any $A,B$ where the above equation is true. Why? Take the limit $z \rightarrow \infty$. The term $\frac{A}{1 - 1.5z^{-1}} + \frac{B}{1 - 0.8z^{-1}}$ factors to maximally $z^{-1}$ in the numerator whereas $\frac{2.6z^{-1} - 1.8z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}} $ contains a $z^{-2}$ term!!!!. So a contradiction !! The correct decomposition is as follows:
$$X_1(z) = \frac{2.6z^{-1} - 1.8z^{-2}}{1 - 2.3z^{-1} + 1.2z^{-2}} =\frac{A}{1 - 1.5z^{-1}} + \frac{B+Cz^{-1}}{1 - 0.8z^{-1}}$$

Taking a common denominator in $\frac{A}{1 - 1.5z^{-1}} + \frac{B}{1 - 0.8z^{-1}}$, equation $(1)$ is now equivalent to
$$2.6z^{-1} - 1.8z^{-2} =A+B + (C -1.5B - 0.8)z^{-1} -1.5Cz^{-1}$$
By identification, we get the following system of equations
$$\begin{bmatrix}
    1.0000 &amp;   1.0000   &amp;      0\\
         0 &amp;  -1.5000  &amp;  1.0000\\
         0  &amp;       0  &amp; -1.5000
\end{bmatrix}
\begin{bmatrix}
A\\B\\C
\end{bmatrix}
=
\begin{bmatrix}
         0\\
    3.4000\\
   -1.8000
\end{bmatrix}
$$
which will give you
\begin{align}
A &amp;= 1.4667\\
B &amp;=  -1.4667 \\
C &amp;= 1.2
\end{align}
Can you continue ?