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    https://b4be22b3d902bfcdbf6f1c96474d...%20Numbers.pdf

    I get 2x^2 -2x+13 with a remainder of 36 when i divide f(x) by x-3

    what can i do to go further?
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    (Original post by timebent)
    https://b4be22b3d902bfcdbf6f1c96474d...%20Numbers.pdf

    I get 2x^2 -2x+13 with a remainder of 36 when i divide f(x) by x-3

    what can i do to go further?
    You shouldn't be getting a remainder.

    Did you do so by long division? You have likely made an error. If you can't spot it feel free to post a picture of your workings for us to look at
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    (Original post by JLegion)
    You shouldn't be getting a remainder.

    Did you do so by long division? You have likely made an error. If you can't spot it feel free to post a picture of your workings for us to look at
    -.-' eugh lemme try again maybe my long division's rusty
    funned up the last bit haha
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    Remainder is 25 not 36.
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    (Original post by B_9710)
    Remainder is 25 not 36.
    no remainders....
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    (Original post by B_9710)
    Remainder is 25 not 36.
    There is no remainder.
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    (Original post by JLegion)
    You shouldn't be getting a remainder.

    Did you do so by long division? You have likely made an error. If you can't spot it feel free to post a picture of your workings for us to look at
    I got as answers x=3 x= \dfrac{1\pm \sqrt 3}{2}
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    (Original post by JLegion)
    There is no remainder.
    (Original post by timebent)
    no remainders....
    I thought the question was divide  2x^2-2x+13 by  x-3 , I can't see the question.
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    (Original post by B_9710)
    I thought the question was divide  2x^2-2x+13 by  x-3 , I can't see the question.
     2x^3 -8x^2 +7x-3 was original question ^^ was what i got the first time i did long division wrong
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    (Original post by timebent)
     2x^3 -8x^2 +7x-3 was original question ^^ was what i got the first time i did long division wrong
    Yes I see now.
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    (Original post by timebent)
    I got as answers x=3 x= \dfrac{1\pm \sqrt 3}{2}
    I think you should check your workings for mistakes; I got x = 3, 1/2 +/- i/2
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    (Original post by timebent)
    I got as answers x=3 x= \dfrac{1\pm \sqrt 3}{2}
    Equation has non real roots.
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    (Original post by JLegion)
    I think you should check your workings for mistakes; I got x = 3, 1/2 +/- i/2
    ok since i can't see where i went wrong

    here's my workings

    (x-3)(2x^2 -2x-1)

     \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}

    \dfrac{2 \pm \sqrt{2^2-4x2x-1}}{4}

     \dfrac{2 \pm \sqrt{12}}{4}

     \dfrac{2 \pm 2\sqrt{3}}{4}

     \dfrac{1 \pm \sqrt{3}}{2}
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    (Original post by timebent)
    ok since i can't see where i went wrong

    here's my workings

    (x-3)(2x^2 -2x-1)

     \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}

    \dfrac{2 \pm \sqrt{2^2-4x2x-1}}{4}

     \dfrac{2 \pm \sqrt{12}}{4}

     \dfrac{2 \pm 2\sqrt{3}}{4}

     \dfrac{1 \pm \sqrt{3}}{2}
    Should get  2x^2-2x+1 as quotient.
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    (Original post by timebent)
    I got as answers x=3 x= \dfrac{1\pm \sqrt 3}{2}
    That's close, sorry I was late to see this.

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    Anyone, is there a faster way?

    It's meant to be part (c)
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    (Original post by Chittesh14)
    Anyone, is there a faster way?

    It's meant to be part (c)
    It's very painful for me to read when you don't use exact values and use decimal approximations.
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    (Original post by B_9710)
    It's very painful for me to read when you don't use exact values and use decimal approximations.
    Hmm.. Don't worry then, I don't think any other way will be much faster.


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    (Original post by Chittesh14)
    Hmm.. Don't worry then, I don't think any other way will be much faster.


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    Probably not.
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    (Original post by B_9710)
    Probably not.
    . Just need to go over chapter 1 before I continue because I don't think I remember everything.


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