Find the coefficient of partial expansion $\frac{2x+3}{(1-x)(1+0.5x+0.5x^2)}$

Question

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 $B$

Answer 1

\begin{equation*} f(x)=\frac{x+\frac{3}{2}}{(x-1)(2+x+x^{2})}=\frac{A}{x-1}+\frac{Bx+C}{% x^{2}+x+2} \end{equation*} cover-up method leads to \begin{equation*} A=\left. f(x)\left( x-1\right) \right\vert _{x=1}=\left. \frac{x+\frac{3}{2}% }{(2+x+x^{2})}\right\vert _{x=1}=\frac{\frac{5}{2}}{4}=\frac{5}{8}. \end{equation*} generalized cover-up method leads to \begin{equation*} Bx+C=\left. f(x)\left( x^{2}+x+2\right) \right\vert _{x^{2}+x=-2}=\left. \frac{x+\frac{3}{2}}{(x-1)}\right\vert _{x^{2}+x=-2} \end{equation*} To find $B$ and $C$, multiply bottom and top of the fraction $\frac{x+\frac{3% }{2}}{(x-1)}$ by $(x-s)$ for some appropriate $s$ such that the denominator (bottom) will be as follows \begin{equation*} (x-1)(x-s)=x^{2}-(s+1)x+s=(x^{2}+x)+s \end{equation*} so we choose \begin{equation*} -(s+1)=+1,\ then\ s=-2 \end{equation*} so \begin{eqnarray*} Bx+C &=&\left. f(x)\left( x^{2}+x+2\right) \right\vert _{x^{2}+x=-2}=\left. \frac{x+\frac{3}{2}}{(x-1)}\right\vert _{x^{2}+x=-2} \\ &=&\left. \frac{\left( x+\frac{3}{2}\right) (x+2)}{(x-1)(x+2)}\right\vert _{x^{2}+x=-2} \\ &=&\left. \frac{x^{2}+2x+\frac{3}{2}x+3}{\left( x^{2}+x\right) -2}% \right\vert _{x^{2}+x=-2} \\ &=&\left. \frac{(x^{2}+x)+\frac{5}{2}x+3}{\left( x^{2}+x\right) -2}% \right\vert _{x^{2}+x=-2},\ replace\ (x^{2}+x)\ by\ -2 \\ &=&\left. \frac{-2+\frac{5}{2}x+3}{-2-2}\right\vert _{x^{2}+x=-2} \\ &=&\frac{\frac{5}{2}x+1}{-4} \\ &=&-\frac{5}{8}x-\frac{1}{4},\ then\ B=-\frac{5}{8},\ and\ C=-\frac{1}{4}, \end{eqnarray*} therefore \begin{equation*} \frac{x+\frac{3}{2}}{(x-1)(2+x+x^{2})}=\frac{5/8}{x-1}+\frac{(-5/8)x+(-1/4)}{% x^{2}+x+2}.\ \ \ \ \blacksquare \end{equation*} EDIT: Actualy, when we replace $x^2+x$ by $-2$ we do not need to know what is the value of $x$ itself!

Answer 2

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

$$C=1+B-A = 1-\frac 54 = -\frac 14$$