Decomposing into partial fractions..

[MKAY03]

I am having trouble with solving this function. The instructions are as follows : Decompose into partial fractions: 3x^2 -6x +8/x(x-2)^2.


This is what I have so far....please let me know if there's anything wrong.

3x^2 -6x+8/x(x-2)^2 = A/x + B/x-2 + C/(x-2)^2
I take x(x-2)^2 and divide by both sides of the function.

So it ends up like this:

3x^2 -6x +8 = A(x-2)^2 + Bx(x-2) + Cx

Then I distribute:
3x^2 -6x+8 = Ax^2 -4Ax +4A + Bx^2 -2Bx +Cx

Combine like terms:

3x^2 -6x +8 = Ax^2 + Bx^2 -4Ax +4A -2Bx + Cx

Then after this is where I get stuck...I know I have to set up a system and solve, but there are 3 different variables so I'm not sure how to do that =/ I know what the final answer is, but I need the missing piece on how to get it!! Please please can anyone help?! Thanks in advance!

 

[arthur ohlsten]

[3x^2-6x+8] / [x[x-2]^2]

I shall determine the constants by a method you might not be familiar with, but it is simple.
[3x^2-6x+8] / {x[x-2]^2} = A/[x-2]^2 + B/[x-2] + c/x

1)
multiply both sides of the equation by x and let x=0
[3x^2-6x+8] / [x-2]^2 = Ax/[x-2]^2+ Bx/[x-2] +C
let x=0
8/4 = C
C=2 answer

2)
multiply both sides by [x-2]^2 and let x=2
[3x^2-6x+8] / x = A +B[x-2] +C[x-2]^2
let x=2
8/2 =A
A=4

3)
there are now two ways to determine the remaining constant. multiply each side by [x-2]^2. Take the derivative with respect to x, and let x=2.

second way
[3x^2-6x+8] / x[x-2]^2 = 4/[x-2]^2 + B /[x-2] + 2
let x=1
5/1 = 4/1 +B/[-1] +2
5=6-B
B=1

Arthur

 

[pitoten]

Another way to get the coefficients...three equations and three unknowns.

3x^2 -6x +8 = Ax^2 + Bx^2 -4Ax +4A -2Bx + Cx


A+B=3

-4A-2B+C=6

4A=8

 

[BigGlenntheHeavy]

\(\displaystyle 3x^2-6x+8/x(x-2)^2\)

\(\displaystyle A. \ \frac{(3x^2-6x+8)(x-2)^2}{x}\)

\(\displaystyle B. \ \frac{3x^2-6x+8}{x*(x-2)^2}\)

\(\displaystyle C. \ 3x^2-6x+\frac{8}{x}(x-2)^2\)

\(\displaystyle Which \ one \ is \ it?\)

\(\displaystyle If \ B, \ then \ (3x^2-6x+8)/(x*(x-2)^2), \ a \ use \ of \ grouping \ symbols \ (a \ common \ courtesy) \ so \ one \ has\)

\(\displaystyle an \ idea \ what \ you \ are \ trying \ to \ get \ across. \\)