partial fractions [c4] helpWatch

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Thread starter 14 years ago
#1
Hey, can someone help me out with this question

18x-2 / (1-x)(1-x)(3+5x) = A/(1-x)(1-x) + B/1-x + C/3+5x

I can get A & C but i dont understand how i can get B, can someone explain it to me please.

to shortcut some of the working out:

18x-2 = A(3+5x) + B(1-x)(3+5x) + C(1-x)(1-x)

When i substitute x=1 for B i get
18(1) - 2 = B(3 + 2 - 5)
0
14 years ago
#2
(Original post by nas7232)
to shortcut some of the working out:

18x-2 = A(3+5x) + B(1-x)(3+5x) + C(1-x)(1-x)

When i substitute x=1 for B i get
18(1) - 2 = B(3 + 2 - 5)
When you substitute x = 1 you get

18-2 = A(3+5) + B(0)

which will only give you A=2 - and I guess you've got C from putting x = -3/5.

To get B you could sub another value like 0 to get:

-2 = 3A + 3B + C

and solve for B (as you already know A and C)

or you could compare coefficients, say of x^2, and get

0 = -5B + C

and again you know C.
0
Thread starter 14 years ago
#3
Thanks 0
14 years ago
#4
Was that supposed to be A/(1-x)^2 ?
0
14 years ago
#5
(18x - 2) / [(1 - x)(1 - x)(3 + 5x)] = A/(1 - x) + B/(1 - x)^2 + C/(3 + 5x) = [A(1 - x)(3 + 5x) + B(3 + 5x) + C(1 - x)^2]/[(1 - x)(1 - x)(3 + 5x)]

Equating numerators:
---> A(1 - x)(3 + 5x) + B(3 + 5x) + C(1 - x)^2 = 18x - 2
Let x = 1:
8B = 16 ---> B = 2
Let x = -3/5:
C(8/5)^2 = -64/5 ---> 64C/25 = -64/5 ---> C = [(25)(-64)]/[(64)(5)] = -5
Let x = 0:
3A + 3B + C = -2 ---> 3A + 6 - 5 = -2 ---> 3A = -3 ---> A = -1

Hence: (18x - 2) / [(1 - x)(1 - x)(3 + 5x)] = 2/(1 - x)^2 - 1/(1 - x) - 5/(3 + 5x)
0
14 years ago
#6
Cheers.
0
Thread starter 14 years ago
#7
(Original post by Nima)
Your welcome. And yes, the thread starter made a mistake with his initial attempt.
Which attempt Sry about that typing error, my method was correct in solving for A & C.
Once again thanks riche for explaining it. Would give you rep, but i gave out too much in 24 hours already. 0
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