Question 394712
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4}{x(x\ -\ 1)(x\ -\ 2)}\ =\ \frac{A}{x}\ +\ \frac{B}{x\ -\ 1}\ +\ \frac{C}{x\ -\ 2}]


Apply the LCD in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4}{x(x\ -\ 1)(x\ -\ 2)}\ =\ \frac{A(x\ -\ 1)(x\ -\ 2)\ +\ Bx(x\ -\ 2)\ +\ Cx(x\ -\ 1)}{x(x\ -\ 1)(x\ -\ 2)}]


Expand the numerator in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4}{x(x\ -\ 1)(x\ -\ 2)}\ =\ \frac{Ax^2\ -\ 3Ax\ +\ 2A\ +\ Bx^2\ -\ 2Bx\ +\ Cx^2\ -\ Cx)}{x(x\ -\ 1)(x\ -\ 2)}]


Combine like terms in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4}{x(x\ -\ 1)(x\ -\ 2)}\ =\ \frac{(A\ +\ B\ +\ C)x^2\ -\ (3A\ +\ 2B +\ C)x\ +\ 2A}{x(x\ -\ 1)(x\ -\ 2)}]


Rewrite the numerator in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{0x^2\ +\ 0x\ +\ 4}{x(x\ -\ 1)(x\ -\ 2)}\ =\ \frac{(A\ +\ B\ +\ C)x^2\ -\ (3A\ +\ 2B +\ C)x\ +\ 2A}{x(x\ -\ 1)(x\ -\ 2)}]


Equate the coefficients:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ +\ B\ +\ C\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -3A\ -\ 2B\ - C\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2A\ =\ 4]


Solve the system for A, B, and C


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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