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Math Help - Partial Fraction Decomposition

  1. #1
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    Partial Fraction Decomposition

    Write the partial fraction decomposition of each rational expression.

    (1) 3x/(x + 2) (x - 4)

    (2) (2x + 4)/(x^3 - 1)
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  2. #2
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    \frac{3x}{(x+2)(x-4)}=\frac{2(x+2)+(x-4)}{(x+2)(x-4)}.
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  3. #3
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    ok but....

    Quote Originally Posted by Krizalid View Post
    \frac{3x}{(x+2)(x-4)}=\frac{2(x+2)+(x-4)}{(x+2)(x-4)}.
    I see that the answer was found but how did you do it?
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  4. #4
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    It's just writting the numerator by using factors of the denominator.

    Or make it as \frac{3x}{(x+2)(x-4)}=\frac{a}{x+2}+\frac{b}{x-4} and find a,b. (I can't provide more help, probably someone else's gonna help a bit more.)
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    It's just writting the numerator by using factors of the denominator.

    Or make it as \frac{3x}{(x+2)(x-4)}=\frac{a}{x+2}+\frac{b}{x-4} and find a,b. (I can't provide more help, probably someone else's gonna help a bit more.)
    This is exactly how to solve the problem. Then multiply both sides by (x+2)(x-4), yielding:

    3x = a(x - 4) + b(x + 2)

    3x = ax - 4a + bx + 2b

    3x = (a+b)x + (2b - 4a)

    from which we deduce

    a + b = 3
    2b - 4a = 0

    and of course the solution is a = 1, b = 2, as Krizalid indicated.
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  6. #6
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    Quote Originally Posted by icemanfan View Post
    This is exactly how to solve the problem. Then multiply both sides by (x+2)(x-4), yielding:

    3x = a(x - 4) + b(x + 2)
    Alternatively, you may just plug in x = 4 and then solve for b. Then plug in x = -2 and solve for a.
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