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Math Help - Irreducible Quadratic Factors (Partial Fractions)

  1. #1
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    Irreducible Quadratic Factors (Partial Fractions)

    Hello, I've been reading the examples in my text over and over and I'm having trouble solving problems with irreducible quadratic factors.

    The problem: Find the constants in the partial fraction decomposition

    (2x-3)/(x-2)((x^2)+4) = A/(x-2) + (Bx+C)/((x^2)+4)

    I solved for A

    2x+4 = A((x^2)+4) + (Bx+C)(x-2)

    Setting x=2, 8 = 8A + 0

    A=1

    Then, like in the example in my text, I distributed the equation and grouped by x's

    2x + 4 = (x^2) + 4 + B(x^2) - 2B + Cx - 2C

    0 = (x^2)(1+B) + x(C-2) - 2(B-C)

    This is where I am lost. I do not know how to solve for B or C.

    Thanks!
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  2. #2
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    Quote Originally Posted by wyhwang7 View Post
    Hello, I've been reading the examples in my text over and over and I'm having trouble solving problems with irreducible quadratic factors.

    The problem: Find the constants in the partial fraction decomposition

    (2x-3)/(x-2)((x^2)+4) = A/(x-2) + (Bx+C)/((x^2)+4)

    I solved for A

    2x+4 = A((x^2)+4) + (Bx+C)(x-2)

    Setting x=2, 8 = 8A + 0

    A=1

    Then, like in the example in my text, I distributed the equation and grouped by x's

    2x + 4 = (x^2) + 4 + B(x^2) - 2Bx + Cx - 2C

    0 = (x^2)(1+B) + x(C-2) - 2(B-C)

    This is where I am lost. I do not know how to solve for B or C.

    Thanks!
    Missed the x there.

    Gives

    2x + 4 = (x^2)(1+B) + x(C-2B) - 2(C-2)

    Comparing coeffts gives B = -1 and C = 0 which works for all 3 powers of x (x^2, x, const) so is correct.
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  3. #3
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    I want the number of x^2 on the right to be the same as on the left.

    So (1+B) = 0 and B = -1

    Then I want 4 = -2(C-2) which gives C = 0.

    Then I check for x: 2 = C - 2B = 0 - (-1)*2 = 2 so it works.
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  4. #4
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    Quote Originally Posted by Thomas154321 View Post
    Missed the x there.

    Gives

    2x + 4 = (x^2)(1+B) + x(C-2B) - 2(C-2)

    Comparing coeffts gives B = -1 and C = 0 which works for all 3 powers of x (x^2, x, const) so is correct.
    Thanks for replying! But I'm confused. How exactly did you find out that B = -1? That would get rid of the x^2, but what about x(C-2B), and how did you derive C = 0? The examples in my text do something like this too, but it doesn't say exactly how these values were derived. Is there a method?
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  5. #5
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    Quote Originally Posted by Thomas154321 View Post
    I want the number of x^2 on the right to be the same as on the left.

    So (1+B) = 0 and B = -1

    Then I want 4 = -2(C-2) which gives C = 0.

    Then I check for x: 2 = C - 2B = 0 - (-1)*2 = 2 so it works.
    Oh, I see! I thought you had to move everything to the right to set the equation to zero.

    Thanks so much!
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