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Partial fraction decomposition exercise 2

  1. Jun 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Hello!
    Here is my second post on the subject partial fraction decomposition. The subject looks pretty easy to learn, but when I try exercises, I do not get to the correct answer. Please, take a look at the exercise below and help me to see my mistakes.

    2. Relevant equations
    (-2x^2 + 20x - 68) / (x^3 + 4x^2 + 4x + 16)

    3. The attempt at a solution
    Step 1: create a form for partial fraction decomposition by factoring the denominator:
    x^3 + 4x^2 + 4x + 16 = (x^2 + 4) ( x + 4)
    x^2 + 4 is in irreducible quadratic form, thus I will work with the above factors
    Step 2: clear denominators
    (-2x^2 + 20x - 68) (x^2 + 4) ( x + 4) / (x^3 + 4x^2 + 4x + 16) = A (x^2 + 4) ( x + 4) / (x + 4) + (Bx + C) (x^2 + 4) ( x + 4) / (x^2 + 4)
    -2x^2 + 20x - 68 = x^2 (A + B) + x ( C + 4B) + 4C + 4A
    Step 3: find values of A, B, C
    A + C = -2
    C + 4B = 20
    4C + 4A = -68 (A + C) = -17
    Matrix A
    1 0 1
    0 1 4
    1 0 1
    Matrix B
    -2
    20
    -17
    And the determinant of matrix A is zero, hence I can't solve the task using Cramer's rule. Obviously, there are mistakes in my approach. What are they? Please, help me to see them.
    Thank you!
     
    Last edited: Jun 25, 2016
  2. jcsd
  3. Jun 25, 2016 #2

    ehild

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    The two sides are not equal. Check if you copied the problem correctly.
     
  4. Jun 25, 2016 #3
    sorry, it's a typo; +16 instead of -68 in the denominator, but all calculations are based on +16 value
     
  5. Jun 25, 2016 #4

    ehild

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    There is a mistake in your equations. The first one is A+B=-2.
     
  6. Jun 25, 2016 #5

    Ray Vickson

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    [tex] \frac{A}{4+x} + \frac{Bx+C}{x^2+4} = \frac{(A+B)x^2 + (4B+C)x +(4A+4C)}{(x+4)(x^2+4)} [/tex]
    Thus, the equations are
    [tex] A+B = -2 \\
    4B+C = 20 \\
    4A+4C=-68
    [/tex]
    Why bother with matrices? It is much easier just to use elementary elimination: the second equation gives ##C = 20-4B##, and putting that into the other two equations gives two linear equations in ##A## and ##B## alone. Solving 2x2 linear systems is pretty easy.
     
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