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Enter (A,B,C) in order below if A, B, and C are the coefficients of the partial fractions expansion of (2x^2 + 3x - 5)/(x(x^2 - 1)) = A/x + B/(x - 1) + C/(x + 1).

 Nov 5, 2020
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\(\frac{2x^2 + 3x - 5}{x(x^2 - 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}\)

multiply both sides by \(x(x^2-1)\):

\(2x^2+3x-5=A(x-1)(x+1)+B(x)(x+1)+C(x)(x-1)\)

Notice that if you set x to equal 0, 1, and -1, respectively, you can cancel out two of the 3 coefficients, which will make it easier to solve. First, set x = 0:

\(-5=-A\\A=5\)

Now set x = 1:

\(2+3-5=2B\\B=0\)

Lastly, set x = -1:

\(2-3-5=2C\\ -6=2C\\C=-3\)

Therefore, the answer is \(\boxed{(5, 0, -3)}\)

 Jul 7, 2021

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