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Partial Fraction Decomposition

Q:  Find the partial fraction decomposition of:

(5x^2 + 3x – 2)/[(x^2)(x + 2)]

A:  Partial fraction what?!?  Our whole life we have been taught how to add 1/4 + 1/6 (get common denominators, blah-blah-blah): 1/4 + 1/6 = 6/24 + 4/24 = 10/24… But, how do you start with 10/24 and realize that you can pull it apart to get 1/4 + 1/6?  That is what partial fraction decomposition is… essentially, it is going backwards. [Long-winded explanation!]

Here is a complicated problem to start us off… This is not a good problem to follow if you are a beginner at partial fractions:

(5x² + 3x – 2)/[(x²)(x + 2)] = A/x + B/x² + C/(x+2)

Now, multiply all parts of the equation by the entire denomiator: (x²)(x + 2)

(x²)(x + 2)(5x² + 3x – 2)/[(x²)(x + 2)] = (x²)(x + 2)A/x + (x²)(x + 2)B/x² + (x²)(x + 2)C/(x+2)

Cancel where possible:

(x²)(x + 2)(5x² + 3x – 2)/[(x²)(x + 2)] = (x²)(x + 2)A/x + (x²)(x + 2)B/ + (x²)(x + 2)C/(x+2)

Therefore:

(1) 5x² + 3x – 2 = A(x)(x + 2) + B(x + 2) + Cx²

Now, pick a “smart” value for x and plug it in.  By “smart”, I mean to pick one that will make things “go away”… For example, x = -2:

5(-2)² + 3(-2) – 2 = A(-2)(-2 + 2) + B(-2 + 2) + C(-2)²

20 – 6 – 2 = A(-2)(0) + B(0) + C(4)

12 = 4C

3 = C

Start back at (1) and plug in C = 3:

5x² + 3x – 2 = A(x)(x + 2) + B(x + 2) + 3x²

Pick another “smart” value for x… Say, x = 0

5(0)² + 3(0) – 2 = A(0)(0 + 2) + B(0 + 2) + 3(0)²

-2 = A(0)(2) + B(2) + 0

-2 = 2B

-1 = B

Start again at (1), plut in B = -1

5x² + 3x – 2 = A(x)(x + 2) + -(x + 2) + 3x²

Now, pick any value for x… I will pick x = 1:

5(1)² + 3(1) – 2 = A(1)(1 + 2) + -(1 + 2) + 3(1)²

5 + 3 – 2 = A(1)(3) – 3 + 3

6 = 3A

2 = A

SO:

(5x² + 3x – 2)/[(x²)(x + 2)] = A/x + B/x² + C/(x+2)

(5x² + 3x – 2)/[(x²)(x + 2)] = 2/x + -1/x² + 3/(x+2)

2 thoughts on “Partial Fraction Decomposition

  1. Ahh ok, that is a nice method, do you know where I went wrong in my matrix though? And once again thanks for the help 🙂

  2. I think where you went wrong was the set up. You had x^3 terms… There should not have been any x^3 terms… I cannot find the original post at the moment, but that is what I remember. It was the matrix, it was the initial set-up

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