SOLUTION: Find partial fraction decomposition of {{{(x^3+1)/(x^2+16)^2}}} It has a repeated quadratic factor in the denominator so I would start it as{{{(x^3+1)/(x^2+16)^2}}}= {{{(Ax+B)/(

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Find partial fraction decomposition of {{{(x^3+1)/(x^2+16)^2}}} It has a repeated quadratic factor in the denominator so I would start it as{{{(x^3+1)/(x^2+16)^2}}}= {{{(Ax+B)/(      Log On


   

Question 1103572: Find partial fraction decomposition of %28x%5E3%2B1%29%2F%28x%5E2%2B16%29%5E2
It has a repeated quadratic factor in the denominator so I would start it as%28x%5E3%2B1%29%2F%28x%5E2%2B16%29%5E2= %28Ax%2BB%29%2F%28x%5E2%2B16%29+%28Cx%2BD%29%2F%28x%5E2%2B16%29%5E2
then you would multiply the three sections by %28x%5E2%2B16%29%5E2
then I got x%5E3%2B1%29=Ax%2BB%28x%5E2%2B16%29%2BCx%2BD
x%5E3%2B1%29=Ax%5E3%2B16Ax%2BBx%5E2%2B16B%2BCx%2BD
Then you would set up a series of equations but from the work above I'm certain it's wrong from where I split up the equation but I dont have a solution to this in my text. Thank you for any help!
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Your work is fine; and the system of equations you need to solve to finish the problem is easily solved. The system is obtained by equating the coefficients of each power on the two sides of the equation:

x^3 term: A = 1
x^2 term: B = 0
x term: 16A+C = 0
constant term: 16B+D = 0

Surely you can finish from there; then the result is easy to check by plugging in the values you found for A, B, C, and D.