<s>[INST] Find the coefficient of partial expansion $\frac{2x+3}{(1-x)(1+0.5x+0.5x^2)}$ I want to decompose the equation:
$$\frac{2x+3}{(1-x)(1+0.5x+0.5x^2)}=\frac{A}{1-x}+\frac{B}{1+0.5x+0.5x^2}$$
I found $A$ by multiple both side with $1-x$ and plug $x=1$. However, it is so difficult to find $B$. Could you help me to find $B$
 [/INST] Starting from Mario G's suggestion in the comments...
$$\frac{x+\frac{3}{2}}{(1-x)(2+x+x^2)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+2}$$
Multiply through by $(x-1)(2+x+x^2)$,
$$x+\frac{3}{2}=A(x^2+x+2)+(Bx+C)(x-1)$$
Collect powers of $x$ on the RHS.
$$x+\frac{3}{2}=(A+B)x^2+(A-B+C)x+(2A-C)$$
Now due to the linear independence of the powers of $x$, we can write three equations, matching the coefficients of each power.
$$\begin{align*}
0&=A+B \\
1&=A-B+C \\
\frac 32&=2A-C
\end{align*}$$
All that's left is to solve the above system. Adding all three equations gives:
$$\frac 52=4A \implies A= \frac 58 $$
Plug this result back into the first equation to get
$$B=-A = -\frac 58$$
And use the second equation to get
$$C=1+B-A = 1-\frac 54 = -\frac 14$$
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