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Math Help - Partial fractions, long division.

  1. #1
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    Partial fractions, long division.

    so the question has two parts and goes like this:

    1) use long division to turn the rational function:

    f(x)=(2x^4-5x^3-9x-12)/(x^3-3x^2+3x-9)

    into the sum of a polynomial and a proper rational function g(x).

    so i did this and got the answer, then i got stuck on part two when it asked:

    2) Look for a small integer root of the denominator of f(x); and use this to find the partial fraction decomposition of g(x)

    help please!? i don't know what it's asking me to do.
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  2. #2
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    Hi Sniperpro

    The question asks you to find the root of (x^3-3x^2+3x-9), then you factorize the denominator. Finally, find the partial fraction decomposition of g(x)
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  3. #3
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    Quote Originally Posted by Sniperpro View Post
    so the question has two parts and goes like this:

    1) use long division to turn the rational function:

    f(x)=(2x^4-5x^3-9x-12)/(x^3-3x^2+3x-9)

    into the sum of a polynomial and a proper rational function g(x).

    so i did this and got the answer, then i got stuck on part two when it asked:

    2) Look for a small integer root of the denominator of f(x); and use this to find the partial fraction decomposition of g(x)

    help please!? i don't know what it's asking me to do.
    and your answer to part (1) was ?
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  4. #4
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    2x+1-[(3x^2+6x-3)/(x^3-3x^2+3x-9)]
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  5. #5
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    i found the root to be x=3, so does this mean i can substitute x^3-3x^2+3x-9 with (x-3)^3?
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  6. #6
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    Hi sniperpro
    Quote Originally Posted by Sniperpro View Post
    i found the root to be x=3, so does this mean i can substitute x^3-3x^2+3x-9 with (x-3)^3?
    No, you can't because (x-3)^3 \neq x^3-3x^2+3x-9. Because you know that x = 3 is the root, you can use Horner's method or long division to factorize x^3-3x^2+3x-9
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  7. #7
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    Quote Originally Posted by Sniperpro View Post
    2x+1-[(3x^2+6x-3)/(x^3-3x^2+3x-9)]
    should be ...

    2x+1 - \frac{3x^2 - 6x + 3}{x^3 - 3x^2 + 3x - 9}

    note that the denominator of the original fraction (and the last term above) equals 0 when x = 3 ... according to the factor theorem, that makes (x-3) a factor ...

    2x+1 - \frac{3x^2 - 6x + 3}{(x-3)(x^2+3)}

    the last question also wants a partial fraction decomposition of the last term ...

    \frac{3x^2 - 6x + 3}{(x-3)(x^2+3)} = \frac{A}{x-3} + \frac{Bx+C}{x^2+3}

    your last task is to find the values of the constants A , B, and C.
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  8. #8
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    long division to factorize?
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  9. #9
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    Quote Originally Posted by Sniperpro View Post
    long division to factorize?
    factorization is just an intermediate process toward the ultimate goal of the partial fraction decomposition.
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  10. #10
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    Hi Sniperpro

    As skeeter said :
    note that the denominator of the original fraction (and the last term above) equals 0 when x = 3 ... according to the factor theorem, that makes (x-3) a factor ...
    So, \frac{x^3 - 3x^2 + 3x - 9}{x-3}=x^2 + 3 and you can factorize x^3 - 3x^2 + 3x - 9.

    x^3 - 3x^2 + 3x - 9 = (x-3)(x^2 + 3) as skeeter has pointed out ^^
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