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The 2010, Honorary Member Welcoming, Undergraduate Degree Results Chat Thread Mk. V watch

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    Hmm, fair enough then...:dontknow:

    This is what Wolfram Alpha says..
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    (Original post by Crazy Paving)
    i got to the point of 1 with remainder -5x - 1 but somehow i dont think thats right :unsure:
    I think you did the same as me, but mixed up signs a bit. Remember its taking away a -ve number so it becomes +ve

    This is so hard and its only the 1st part of the method :emo:
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    (Original post by maxfire)
    Cos if it was +2...

    \dfrac{x^2-2x-3}{x^2+3x+2} = \dfrac{(x-3)(x+1)}{(x+2)(x+1)} = \dfrac{x-3}{x+2}

    Having -2 at the end changes it a lot, and I doubt you'd get something like that in C4.
    Improper partial fractions...

    That link did help, thanks.

    Anon, you did get it right :yep:, how did you do it?
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    (Original post by Loz17)
    I think you did the same as me, but mixed up signs a bit. Remember its taking away a -ve number so it becomes +ve

    This is so hard and its only the 1st part of the method :emo:
    my signs seem right unless you divided them the wrong way round :dontknow:

    which one is which?
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    (Original post by Loz17)
    Improper partial fractions...

    That link did help, thanks.

    Anon, you did get it right :yep:, how did you do it?
    :jive:

    erm, i think you must have divided them the wrong way round then
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    (Original post by Crazy Paving)
    :love:
    :ahee:
    Ach! This has just reminded me I have maths homework to do
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    (Original post by Crazy Paving)
    :jive:

    erm, i think you must have divided them the wrong way round then
    Maybe...

    I divided the x^2 terms (I can't show the full method on latex I don't think) to get the 1, then I multiplied all of the divisor by one, and took them away from each other
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    (Original post by Loz17)
    Maybe...

    I divided the x^2 terms (I can't show the full method on latex I don't think) to get the 1, then I multiplied all of the divisor by one, and took them away from each other
    could you do me a favour please because somethings been bugging me for a while.

    Spoiler:
    Show
    On the OP, can you take the Re: part out of the title? It shouldn't be there. :emo: Thanks.


    erm yeah its hard to show on here :erm:
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    I've finished all my hw :jiggy:
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    (Original post by Smeh)
    :ahee:
    Ach! This has just reminded me I have maths homework to do
    I always have maths homework to do.
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    :hand:
    • Thread Starter
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    Cba to upload to Photobucket

    My method is in attachment. Sorry its not clear
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    (Original post by Loz17)
    Cba to upload to Photobucket

    My method is in attachment. Sorry its not clear
    yeah you divided them the wrong way round
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    (Original post by Loz17)
    Cba to upload to Photobucket

    My method is in attachment. Sorry its not clear
    lmao paint!

    Looks right though.

    Oh wait.

    • Thread Starter
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    (Original post by Crazy Paving)
    yeah you divided them the wrong way round
    :hmmmm:

    The example in the book said to divide the demoninator by the numerator
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    (Original post by Loz17)
    :hmmmm:

    The example in the book said to divide the demoninator by the numerator
    when you told us the question, you said it the other way round :hmmmm:

    you're suppose to divide the numerator by the denominator i thought
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    I've confused myself and others

    My question is....

    \frac{x^2+3x-2}{(x+1)(x-3)} \Rightarrow \frac{x^2+3x+2}{x^2-2x-3}
    Then my method above in paint follows that
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    oh im confused now because then that would come out with your answer of 1 r 5x + 1
    • Thread Starter
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    I got it, I tried both answers and it was 5x + 1 in the end.

    Thanks for the help everyone! :hugs:
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    yeeeeah let's :party:
 
 
 
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