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OpenStudy (anonymous):

partial fraction decomposition 5x^2-9x+19 over (x-4)(x^2+5) can someone help?

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OpenStudy (anonymous):

I can :) You have a linear factor (x-4) and an irreducible quadratic (x^2+5) So you should look for something in the form: \[\frac{5x^2-9x+19}{(x-4)(x^2+5)}=\frac{A}{(x-4)}+\frac{Bx+C}{(x^2+5)}\] Get a common denominator: \[\frac{A(x^2+5)+(x-4)(Bx+C)}{(x-4)(x^2+5)}\] From here set the numerators equal. \[5x^2-9x+19=Ax^2+5A+Bx^2+Cx-4Bx-4C\] From here you know that: A+B=5; 5A-4C=19; C-4B=-9 Solving you get that: A=3,B=2, C=-1 So you get:\[\frac{3}{(x-4)}+\frac{2x-1}{(x^2+5)}\]

OpenStudy (anonymous):

Sorry it took so long to type up.

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