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partial fraction decomposition 3x-4/x^2-4x-12
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You have: \[\frac{3x-4}{x^2-4x-12}\] Factor the bottom: \[\frac{3x-4}{(x-6)(x+2)}\] You have two linear factors. So you are looking for something in the form of: \[\frac{A}{x-6}+\frac{B}{x+2}\] So get a common denominator and set the numerators equal giving: \[3x-4=A(x+2)+B(x-6)\] Multiply and collect like terms: \[3x-4=Ax+Bx+2A-6B\] So you know that A+B=3 2A-6B=-4 B=3-A 2A-6(3-A)=-4 2A-18+6A=-4 8A=14 A=7/4 So B=3-(7/4)=5/4 So your decomposition is: \[\frac{7}{4(x-6)}+\frac{5}{4(x+2)}\]
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