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Math Help - Finding the ordered Triplet

  1. #1
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    Exclamation Finding the ordered Triplet

    Find the ordered triplet (A,B,C) if:


    (6x^2 - 8x -20) / [(x+2)(x-3)(x-2)] = (A/(x+2)) + (B/(x-3)) + (C/(x-2))


    **hint: mulitply both sides by (x+2)(x-3)(x-2)
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  2. #2
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    Quote Originally Posted by Mr_Green View Post
    Find the ordered triplet (A,B,C) if:


    (6x^2 - 8x -20) / [(x+2)(x-3)(x-2)] = (A/(x+2)) + (B/(x-3)) + (C/(x-2))


    **hint: mulitply both sides by (x+2)(x-3)(x-2)
    Multiply through and you get,

    6x^2-8x-20=A(x-3)(x-2)+B(x+2)(x-2)+C(x+2)(x-3)
    Now this is a partial fraction decomposition which can always be achived. Thus this is true for all x.

    If x=2
    6(2)^2-8(2)-20=A(2-3)(0)+B(2+2)(0)+C(2)(-1) you can solve for C

    If x=3
    6(3)^3-8(3)-20=A(0)(3-2)+B(3+2)(3-2)+C(5)(0) you can solve for B

    If x=-2
    6(-2)^2-8(-2)-20=A(-5)(-4)+B(0)(-2-2)+C(0)(-5) you can solve for A
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  3. #3
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    what i got.

    i got

    A = 1

    B = 23.6

    C = 6

    Could someone check that i got this right?
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  4. #4
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    Nvm, i got the answer and checked it...

    A = 1

    B = 2

    C = 3

    thanks everyone!
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  5. #5
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    Quote Originally Posted by Mr_Green View Post
    i got
    A = 1
    B = 23.6
    C = 6
    Could someone check that i got this right?
    Hello, Mr Green,

    TPH had made a small but powerful typo:
    In

    If x=3
    6(3)^3-8(3)-20=A(0)(3-2)+B(3+2)(3-2)+C(5)(0) you can solve for B
    the exponent of the first 3 should be 2

    Then you'll get: B = 2

    EB
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  6. #6
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    factoring

    thanks
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