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Math Help - Basic Algebra?

  1. #1
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    Basic Algebra?

    Problem
    \int\frac{10x-2x^2}{(x-1)^2(x+3)}
    I did all the steps correctly until this part

    A+C = -2
    2A+ B -2C = 10
    -3A +3B + C = 0

    Attempt:
    I said C = -A-2
    B = -2A +2C + 10

    So
    -3A + 3(2A + 2(-A-2) + 10) + (-A-2) = 0
    -3A + 3(2A -2A -4 + 10) -A-2 = 0
    -3A + -12 +30 -A - 2 = 0
    -4A +16 = 0
    -4A = -16

    Answer:
    A = 4


    Correct Answer
    A= 1, B= 2, C= -3

    *This is related to partial fractions in Calculus.
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  2. #2
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    Hi

    So
    -3A + 3(-2A + 2(-A-2) + 10) + (-A-2) = 0
    -3A + 3(-2A -2A -4 + 10) -A-2 = 0
    -3A + -12A -12 +30 -A - 2 = 0
    -16A +16 = 0
    -16A = -16

    Answer:
    A = 1
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Fallen186 View Post
    Problem
    \int\frac{10x-2x^2}{(x-1)^2(x+3)}
    I did all the steps correctly until this part

    A+C = -2
    2A+ B -2C = 10
    -3A +3B + C = 0

    Attempt:
    I said C = -A-2
    B = -2A +2C + 10

    So
    -3A + 3(-2A + 2(-A-2) + 10) + (-A-2) = 0
    -3A + 3(-2A -2A -4 + 10) -A-2 = 0
    -3A + -12 +30 -A - 2 = 0
    -4A +16 = 0
    -4A = -16

    Answer:
    A = 4


    Correct Answer
    A= 1, B= 2, C= -3

    *This is related to partial fractions in Calculus.
    You missed a sign at the beginning (See chage in red). See if that fixes it. I didn't bother to work it out for you.
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  4. #4
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    Quote Originally Posted by Fallen186 View Post
    Problem
    \int\frac{10x-2x^2}{(x-1)^2(x+3)}

    Correct Answer
    A= 1, B= 2, C= -3

    *This is related to partial fractions in Calculus.

    \frac{10x-2x^2}{(x-1)^2(x+3)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+3}<br />

    10x-2x^2 = A(x-1)(x+3) + B(x+3) + C(x-1)^2

    x = 1 ... 4B = 8 ... B = 2

    x = -3 ... 16C = -48 ... C = -3

    x = 0 ... -3A + 3B + C = 0 ... -3A + 6 - 3 = 0 ... A = 1
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