AllExperts > Advanced Math 
Search      
Advanced Math
Volunteer
Answers to thousands of questions
 Home · More Advanced Math Questions · Answer Library  · Encyclopedia ·
More Advanced Math Answers
Question Library

Ask a question about Advanced Math
Volunteer
Experts of the Month
Expert Login

Awards

About Us
Tell friends
Link to Us
Disclaimer

 
 
 
 
About Steve Holleran
Expertise
I can help with all math questions from basic math to Calculus. Whether it`s consumer questions, or questions from high school or college students, I have probably dealt with it at some time in my career.

Experience
33 years teaching experience in NJ public schools

Education/Credentials
B.S. Mathematics : Wake Forest University 1972 M.S. Mathematics : Monmouth University 1981

 
   

You are here:  Experts > Science > Mathematics > Advanced Math > Partial fraction

Advanced Math - Partial fraction


Expert: Steve Holleran - 10/2/2008

Question
Express the following in partial fraction:


(x^2-x)/((x^2+3)(x^2+2))


i have tried it many times for hours, again & again still cant get it right. Much appreciated.

Answer
Hello,

I think it goes like this:


 (x^2-x)/(x^2+3)(x^2+2) =(Ax + B)/(x^2 + 3)+(Cx + D)/(x^2+2)

          x^2 - x = (Ax + B)(x^2+2) + (Cx + D)(x^2 + 3)

     x^2 - x = Ax^3 + Bx^2 + 2Ax + 2B + Cx^3 + Dx^2 + 3Cx + 3D

     x^2 - x = x^3(A + C) + x^2(B + D) + x(2A + 3C) + (2B + 3D)

Then  A + C = 0      B+ D = 1    2A - 3C = -1     2B + 3D = 0

Solving the A,C system:

     2A - 3C = -1   ------->   2A - 3C =-1

      A +  C = 0    ----->    -2A -2C = 0
                             -------------

                                 -5C = -1  so C = 1/5  ,then A = -1/5

If you solve the B, D system, you get B = 3 and D = -2

So, the solution is

   (-1/5 x + 3)/(x^2 + 3)   + (1/5 x - 2)/(x^2 + 2)


Hope this helps
Steve

Add to this Answer   Ask a Question


 
User Agreement | Privacy Policy | Kids' Privacy Policy | Help
Copyright  © 2008 About, Inc. AllExperts, AllExperts.com, and About.com are registered trademarks of About, Inc. All rights reserved.