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Math Help - Algebraic Fractions question

  1. #1
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    Algebraic Fractions question

    I'v been stuck on this question and I dont know how to solve it. I have to use the remainder theorem to find the answer for (4x - 5x + 3x - 14) / (x + 2x - 1).

    Since I can't expand (x + 2x - 1) so I cannot find any values for "x". Can somebody please help I have to hand this in after tomorrow.

    Thanks
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  2. #2
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    Quote Originally Posted by neo658 View Post
    I'v been stuck on this question and I dont know how to solve it. I have to use the remainder theorem to find the answer for (4x³ - 5x² + 3x - 14) / (x² + 2x - 1).

    Since I can't expand (x² + 2x - 1) so I cannot find any values for "x". Can somebody please help I have to hand this in after tomorrow.

    Thanks
    Code:
     4x^3-5x^2+3x-14 -:- x^2+2x-1=4x-13
    -4x^3-8x^2+4x
    ------------------
    -13x^2+ 7x  -14
    +13x^2+26x-13
    --------------------
    33x-27 (this is the remainder)
    This is mine 27th Post!!!
    Last edited by ThePerfectHacker; October 3rd 2006 at 10:44 AM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Code:
     4x^3-5x^2+3x-14 -:- x^2+2x-1=4x-13
    -4x^3-8x^2+4x
    ------------------
    -13x^2+ 7x  -14
    +13x^2+26x-13
    --------------------
    33x-27 (this is the remainder)
    This is mine 27th Post!!!
    To expand on TPH's answer a little bit it is based on the remainder theorem:
    Given polynomials a and b, if the remainder of a/b is r, then a = b*q + r, where q and r are polynomials and the degree of r is 0 <= r < b.

    Here a = 4x^3-5x^2+3x-14, b = x^2+2x-1, q = 4x + 13, and r = 33x - 27.

    -Dan
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    Quote Originally Posted by topsquark View Post
    To expand on TPH's answer a little bit it is based on the remainder theorem:
    Given polynomials a and b, if the remainder of a/b is r, then a = b*q + r, where q and r are polynomials and the degree of r is 0 <= r < b.

    Here a = 4x^3-5x^2+3x-14, b = x^2+2x-1, q = 4x + 13, and r = 33x - 27.
    Let me say it even better.
    Given polynomials f(x) and g(x) then there exists unique polynomials q(x) and r(x) with deg (r)<deg(g) such that,
    f(x)=q(x)f(x)+r(x)

    [Existence/Uniqueness happens to be of paramount importance in field theory]
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  5. #5
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    uh guys can ya please make this like simpler I dint understandu guys at all
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  6. #6
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    Quote Originally Posted by neo658 View Post
    uh guys can ya please make this like simpler I dint understandu guys at all
    topsquark is to blame.
    ---
    Look at my first post.
    Look at how I did long division, and understand it.
    The number on the bottom which is left you cannot divide futher.
    That number is the "remainder".
    The number after the equal sign is the "quotient".
    Now you have the quotient and remainder after division.
    Attached Thumbnails Attached Thumbnails Algebraic Fractions question-picture10.gif  
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  7. #7
    Forum Admin topsquark's Avatar
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    I'm sowwy!

    -Dan
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  8. #8
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    thanks loads guys now so much clearer =)
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