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Find all values of A, B, C and C such that: $$ \frac{x-1}{(x-1)(x-2)(x-2)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{(x-2)^2} $$

I keep getting into a loop in which: $$ x - 1 = Ax^2 - 4Ax + 4A + Bx^2 - Bx - 2B + Cx + C $$

Which then creates a system which will create equal versions of existing equations in the system. $$ -1 = 4A - 3B + C $$ $$ -1 = -Ax - Bc + 4A + B - C $$

Are there any methods of approaching this?

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    $\begingroup$ Isn’t it just $A=B=0$, $C=1$ by looking at it? $\endgroup$
    – Aig
    Commented Mar 1 at 3:21
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    $\begingroup$ A, B, C and C? Could you also check the signs as if this is an exercise in partial fractions, something like $\frac{x-1}{(x+1)(x-2)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$ might be a more typical set up $\endgroup$
    – Henry
    Commented Mar 1 at 3:28
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    $\begingroup$ @Durian: Why did you not cancel $(x-1)$ in num/den ? $\endgroup$
    – Narasimham
    Commented Mar 1 at 3:39
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    $\begingroup$ Your system of equations looks totally wrong, based on the previous line. Also you replace $x+1$ with $x-1$ and $x-2$ with $x+2$. I'd try redoing this whole thing. $\endgroup$
    – pancini
    Commented Mar 1 at 3:47
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    $\begingroup$ You got into the "loop": $x-1 = A(x-2)^2 + B(x-2)(x+1) + C(x+1)$, but how? $\endgroup$
    – peterwhy
    Commented Mar 1 at 3:53

2 Answers 2

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Mistake in The Problem / Your Post

First, there are mistakes in the decomposition. The proper equation should be:

$$ \frac{x-1}{(x-1)(x-2)^{2}} = \frac{A}{x-1}+ \frac{B}{x-2}+ \frac{C}{(x-2)^{2}} $$

from which we have

$$ \frac{x-1}{(x-1)(x-2)^{2}} = \frac{A(x-2)^{2}+B(x-1)(x-2)+C(x-1)}{(x-1)(x-2)^{2}} $$

Comparing The Numerators

Because the denominators on the LHS and RHS are equal, the numerators must be equal as well:

$$ x-1 = (A+B)x^{2}+(-4A-3B+C)x+(4A+2B-C) $$

which can be written in matrix-vector notations:

$$ \begin{Bmatrix} x^{2} \\ x \\ 1 \end{Bmatrix}^{\top} \begin{Bmatrix} 0 \\ 1 \\ -1 \end{Bmatrix} = \begin{Bmatrix} x^{2} \\ x \\ 1 \end{Bmatrix}^{\top} \begin{bmatrix} 1 & 1 & 0 \\ -4 & -3 & 1 \\ 4 & 2 & -1 \end{bmatrix} \begin{Bmatrix} A \\ B \\ C \end{Bmatrix} $$

and the solution is

$$ \begin{Bmatrix} A \\ B \\ C \end{Bmatrix} = \begin{bmatrix} 1 & 1 & 0 \\ -4 & -3 & 1 \\ 4 & 2 & -1 \end{bmatrix}^{-1} \begin{Bmatrix} 0 \\ 1 \\ -1 \end{Bmatrix} $$

i.e. $A=B=0$ and $C=1$

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Another point of view like below: $$* \ \ \frac{x-1}{(x+1)(x-2)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$$ multilply * by $(x+1)$ then put $x=-1$ $$(x+1) \times (\frac{x-1}{(x+1)(x-2)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}) \\ \frac{x-1}{(x-2)(x-2)} = \frac{A}{1} + (x+1)(\frac{B}{x-2} + \frac{C}{(x-2)^2}) \\x=-1 \to \frac {-1-1}{(-1-2)^2}=A+0 \to A=\frac {-2}{9}$$ then multiply * by $(x-2)^2$ to finc $C$ $$(x-2)^2(\frac{x+1}{(x-1)(x-2)(x-2)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{(x-2)^2})\\\frac {x-1}{x+1}=(x-2)^2(\frac{A}{x+1} + \frac{B}{x-2})+C \\x=2 \to \frac {2-1}{2+1}=0+C$$now to find B you can put any $x\neq -1,2$ forexample $x=0$ $$\frac{x-1}{(x+1)(x-2)(x-2)} = \frac{\frac {-2}{9}}{x+1} + \frac{B}{x-2} + \frac{\frac {1}{3}}{(x-2)^2} \\x=0 \to \frac {0-1}{(0+1)(0-2)^2}=\frac{\frac {-2}{9}}{0+1} + \frac{B}{0-2} + \frac{\frac {1}{3}}{(0-2)^2}$$

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