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Mathematics 19 Online
OpenStudy (katieb):

What is the partial fraction decomposition? Please help me. These are impossible. :( -5x-15/(x+5)^2

OpenStudy (anonymous):

You have a repeated linear factor in the denominator, so you decompose the fraction as such:\[\frac{-5x-15}{(x+5)^2}=\frac{A}{x+5}+\frac{B}{(x+5)^2}\]You then multiply both sides by (x+5)^2 to obtain\[-5x-15=A(x+5)+B\]Expand and collect like terms:\[-5x-15=Ax+(5A+B)\]In order for this to be true, we must have\[A=-5\](identifying it with the coefficient of x on the left) and \[5A+B=-15\]Since A has to be -5, we may solve for B:\[-15=5A+B=5(-5)+B=-25+B \rightarrow B=-15+25=10\]So A = -5 and B = 10, and your decomposition is\[\frac{-5x-15}{(x+5)^2}=-\frac{5}{x+5}+\frac{10}{(x+5)^2}\]

OpenStudy (anonymous):

You can always check by cross-multiplying the right-hand side to see if it gets you back to what you had originally.

OpenStudy (katieb):

it's saying that's incorrect. :(

OpenStudy (anonymous):

Sorry, Katie, but the answer is correct. I've double-checked. Maybe you've entered something wrong?

OpenStudy (katieb):

Ok, you're a genius. I forgot the negative. Please help me with 2 more problems?

OpenStudy (katieb):

Done.

OpenStudy (katieb):

partial decomposition of... x^3+x^2-5/(x^2+7)^2

OpenStudy (anonymous):

OpenStudy (katieb):

What is the partial fraction decomposition? 8x^2+20x+77/x(x-9)(x-8)

OpenStudy (katieb):

Last one please and thank you so much, but it's due in 8 minutes.

OpenStudy (anonymous):

I can give you what I got, but I've rushed and can't guarantee. \[\frac{77/72}{x}+\frac{60535/616}{x-9}-\frac{63299/693}{x-8}\]

OpenStudy (anonymous):

wait

OpenStudy (anonymous):

-749/8 for the x-8 one 905/9 for x-9 one 77/72 for x one

OpenStudy (katieb):

aw man too late

OpenStudy (katieb):

thank you so much for trying! you're awesome!

OpenStudy (anonymous):

oh well...next time, don't leave it to the last minute! :p

OpenStudy (anonymous):

Go to khanacademy.org or YouTube if you need to see how partial fraction decomposition is done :)

OpenStudy (katieb):

thank you!

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