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adk2
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#1
Report Thread starter 15 years ago
#1
Put (2-3x)/(1-3x+2x^2) into partial fractions.

Please advise.
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KingAS
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#2
Report 15 years ago
#2
factorise the denominator to (x-1)(2x-1), then split into partial fractions
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username9816
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#3
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#3
(Original post by adk2)
Put (2-3x)/(1-3x+2x^2) into partial fractions.

Please advise.
(2 - 3x)/(1 - 3x + 2x^2) = (2 - 3x)/[(2x - 1)(x - 1)] = A/(2x - 1) + B/(x - 1) = [A(x - 1) + B(2x - 1)]/[(2x - 1)(x - 1)]

Where A and B are constants to be determined.

Equating numerators gives:
---> A(x - 1) + B(2x - 1) = 2 - 3x

Let x = 1:
---> B = -1

Let x = 2:
---> A + 3B = -4
---> A = -4 + 3
---> A = -1

Hence: (2 - 3x)/(1 - 3x + 2x^2) = -[1/(2x - 1) + 1/(x - 1)]
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