SOLUTION: IN THE PARTIAL FRACTION DECOMPOSITION of x^2/[(x-1)^2 (x+1)^2] the format is as follows: A/(x-1) + B/(x-1)^2 +C/(x+1) + D/(x+1)^2 I let x = 1 and got B= 1/4 (which is correc

Algebra ->  Matrices-and-determiminant -> SOLUTION: IN THE PARTIAL FRACTION DECOMPOSITION of x^2/[(x-1)^2 (x+1)^2] the format is as follows: A/(x-1) + B/(x-1)^2 +C/(x+1) + D/(x+1)^2 I let x = 1 and got B= 1/4 (which is correc      Log On


   

Question 1133349: IN THE PARTIAL FRACTION DECOMPOSITION of x^2/[(x-1)^2 (x+1)^2] the format
is as follows:
A/(x-1) + B/(x-1)^2 +C/(x+1) + D/(x+1)^2
I let x = 1 and got B= 1/4 (which is correct)
THEN let x=-1 to get D= 1/4
Am stuck at this point Can not get A or C (A = 1/4 and C = -1/4 but how to get this?
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
IN THE PARTIAL FRACTION DECOMPOSITION of x^2/(x-1)^2 (x+1)^2 the format is as follows:

Multiply through by the LCD:

I let x = 1




+1+=+B%2A%282%29%5E2

+1+=+B%2A4

1%2F4=B

and got B= 1/4 (which is correct) THEN let x=-1 to get D= 1/4




+1+=+D%2A%28-2%29%5E2

+1+=+D%2A4

1%2F4=D

Am stuck at this point Can not get A or C
Go back to:



Substitute the values you got for B and D



Clear of fractions:



Substitute some number you haven't substituted before. 

(Note: You don't have to just substitute numbers that make terms equal to zero.
You can substitute ANY number for x and it will always give a true equation,
even if none of the terms become 0. So the easiest number you haven't
substituted is x=0.


+0+=+4A%2A%28-1%29%281%29%5E2+%2B+%281%29%5E2+%2B4C%2A%281%29%28-1%29%5E2+%2B+%28-1%29%5E2
+0+=+-4A+%2B+1+%2B4C+%2B+1
+0+=+-4A+%2B+4C+%2B+2
4A-4C=2
2A-2C=1

The next easiest number you haven't substituted is x=2


+4%2A4+=+4A%2A%281%29%283%29%5E2+%2B+%283%29%5E2+%2B4C%2A%283%29%281%29%5E2+%2B+%281%29%5E2
+16+=+4A%2A%289%29+%2B+9+%2B4%2AC%2A%283%29+%2B+1
+16+=+36A+%2B+10+%2B12C
+6+=+36A+%2B+12C
+1+=+6A+%2B+2C
+6A+%2B+2C+=1

Solve the system of equations:

system%282A-2C=1%2C6A+%2B+2C+=1%29

Get A=1%2F4, C=-1%2F4

That's how you do it. If you run out of numbers to substitute for x that
cause terms to be 0, substitute some numbers that DON'T cause any terms
to be 0, and solve a system of equations.  Any numbers will do.  I just
picked the next easiest numbers x=0 and x=2. 

Edwin