Partial Fraction Decomposition! (Confirm my answers please)

[Kevinmon]

I have to write out the decomposition and find the numerical values of A, B, C, etc.

Problem:
x^2+9x-12
(3x-1)(x+6)^2

My work so far:
A/(3x-1) + B/(x+6) + C/(x+6)^2

x^2+9x-12 = A(x+6)^2 + B(3x-1)(x+6) + C(3x-1)

And so I solved for A, B, and C but got really weird fractions like 147/361.

If you could help out here that would be great! Thanks!

 

[Deleted member 4993]

Re: Partial Fraction Decomposition!

I have to write out the decomposition and find the numerical values of A, B, C, etc.

Problem:
x^2+9x-12
(3x-1)(x+6)^2

My work so far:
A/(3x-1) + B/(x+6) + C/(x+6)^2

x^2+9x-12 = A(x+6)^2 + B(3x-1)(x+6) + C(3x-1)

And so I solved for A, B, and C but got really weird fractions like 147/361.

If you could help out here that would be great! Thanks!

I get A = - 80/361

There are no guarantees that those coefficients will be nice integers. In a "real world" problem - almost certainly those would be "wierd" numbers.

 

[Kevinmon]

Here's something to shoot for:

\(\displaystyle \L\\\frac{-26}{361(3x-1)}+\frac{129}{361(x+6)}-\frac{33}{19(x+6)^{2}}\)

I get:
A=-80/361
B=147/361
C=30/19

Eh?

 

[galactus]

You're very right. I made a typo in the original and that's all it took to through me off.

 

[Deleted member 4993]

(x^2 + 9x -12)/[(3x-1)(x+6)^2] = A/(3x-1) + B/(x+6) + C/(x+6)^2

multiply both sides bu (3x-1)

(x^2 + 9x -12)/(x+6)^2 = A + B(3x-1)/(x+6) + C(3x-1)/(x+6)^2

Take limit of both sides x -> 1/3

(-80/9)/(361/9) = A

A = - 80/361

Similarly you can find 'C' by multiplying both sides by (x+6)^2 and taking limit as x -> -6

(36-54-12)/(-18-1) = C

C = 30/19