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Math Help - polynomial division, remainder

  1. #1
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    polynomial division, remainder

    I need to find the remainder of \frac{x^7-6x^2+3}{x-1}

    Long division

    ------- x^6 +x^5 + x^4 + x^3 + x^2 + 5x + 5

    __________________________________________________ ____
     x-1 [ x^7 +0x^6 + 0x^5 +0x^4+0x^3 + 6x^2+0x +3 ]
    .....  -(x^7 -x^6)
    ................ x^6 + 0x^5
    .............. -(x^6-x^5)
    ......................... x^5+0x^4
    ....................... -(x^5-x^4)
    ................................. x^4+0x^3
    ...................................... -(x^3-x^2)
    ......................................... x^3-6x^2
    ................................................ 5x^2+0x
    .............................................. -(5x^2-5x)
    .................................................. ....... 5x+3
    .................................................. ..... -(5x-5)
    .................................................. .................8

    Synthetic Division

    1 [ 1 0 0 0 0 -6 3]
    ........1 1 1 1 1 5
    ........1 1 1 1 5 8

    It seems that the remainder is \frac{8}{x-1}, however I know the answer is P(1)=1-6+3=2. Where am I going wrong here? Why does P(1) give the remainder?
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  2. #2
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    Quote Originally Posted by adkinsjr View Post

    __________________________________________________ ____
     x-1 [ x^7 +0x^6 + 0x^5 +0x^4+0x^3 + 6x^2+0x +3 ]
    .....  -(x^7 -x^6)
    ................ x^6 + 0x^5
    .............. -(x^6-x^5)
    ......................... x^5+0x^4
    ....................... -(x^5-x^4)
    ................................. x^4+0x^3

    ...................................... -(x^3-x^2)
    ......................................... x^3-6x^2
    ................................................ 5x^2+0x


    .............................................. -(5x^2-5x)
    .................................................. ....... 5x+3
    .................................................. ..... -(5x-5)
    .................................................. .................8
    I don't know if this will get you to your desired answer, but I do see a mistake in your work where I bolded above. Notice in the first bolded line you have minus a negative x^2. So you should have x^2-6x^2=-5x, not positive 5x.
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  3. #3
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    Quote Originally Posted by adkinsjr View Post
    I need to find the remainder of \frac{x^7-6x^2+3}{x-1}

    Long division

    ------- x^6 +x^5 + x^4 + x^3 + x^2 + 5x + 5

    __________________________________________________ ____
     x-1 [ x^7 +0x^6 + 0x^5 +0x^4+0x^3 + 6x^2+0x +3 ]
    .....  -(x^7 -x^6)
    ................ x^6 + 0x^5
    .............. -(x^6-x^5)
    ......................... x^5+0x^4
    ....................... -(x^5-x^4)
    ................................. x^4+0x^3
    ...................................... -(x^3-x^2)
    ......................................... x^3-6x^2
    ................................................ 5x^2+0x
    .............................................. -(5x^2-5x)
    .................................................. ....... 5x+3
    .................................................. ..... -(5x-5)
    .................................................. .................8

    Synthetic Division

    1 [ 1 0 0 0 0 -6 3]
    ........1 1 1 1 1 5
    ........1 1 1 1 5 8

    It seems that the remainder is \frac{8}{x-1}, however I know the answer is P(1)=1-6+3=2. Where am I going wrong here? Why does P(1) give the remainder?
    Read this: Polynomial remainder theorem - Wikipedia, the free encyclopedia
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