Printable View
I know how to do these problems but this one is giving me some trouble:
Expand into partial fractions
5/(x+1)(x^2-1)
***Should I break it down into:
1) [A/(x+1)]+[B/(x-1)]+[C/(x+1)] or
2) [A/(x+1)]+[B/(x+1)^2]+[C/(x-1)]
I tried it both ways and came up with different possibilities for the constraints:
If Broken Down Like #1 from Above
A+B+C=0
2B = 0
A+B-C=5
If Broken Down Like #2 from Above
A+C=0
B+2C=0
A+B-C=5
Which way gives you the LCD that is the denominator of the original expression?
I'm not sure because you can break down the original denominator (x+1)(x^2-1) into (x+1)(x+1)(x-1)
Yes, and if you were to combine the 3 terms of both the choices you listed, which one would give you the correct common denominator?
Option 1 would give me that..right?
No it wouldn't. Since two of them haveas the denominator, the LCD is
, NOT
as required.
When you have repeated factors, you need one denominator to bein your partial fraction.
so when I write it in the partial fractions I write it as:
[A/(x+1)]+[B/(x+1)^2]+[C/(x-1)]
??
Yes
5/(x+1)(x^2-1) = A/(x+1) + B/(x+1)^2 + C/(x-1)
multiply by LCD to get:
5= A(x+1)(x-1) + B(x-1) + C(x+1)^2
5= A(x^2-1) + B(x-1) + C(x^2+2x+1)
5= Ax^2 - A + Bx - B + Cx^2 + 2Cx + C
Group Common Terms:
5= x^2(A+C) + x(B+2C) + (-A-B+C)
Constraints:
A+C=0
B+2C=0
A+B-C=5
This all looks fine. Now solve for A,B,C.Quote:
Originally Posted by rrooggeerr
I get:
A= -5/4
B= -5/2
C= 5/4
I know something is wrong though because it doesn't make the third constraint true :(
never mind, i got it!!
A = 5/4
B= 5/2
2= -5/4
THANK YOU!
There is an alternate method for partial fraction decomposition you may be interested in:
Heaviside cover-up method - Wikipedia, the free encyclopedia