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Algebraic Long division help! Watch

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    could anyone tell me where i went wrong please?

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    P.S. sorry i don't know why it's upside down :/
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    (Original post by Prista)
    could anyone tell me where i went wrong please?

    Name:  photo.jpg
Views: 734
Size:  341.4 KB

    P.S. sorry i don't know why it's upside down :/
    Type out the question, I have another method instead of long division, its the bracket method so much easier and quicker, but needs practice
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    (Original post by Prista)
    could anyone tell me where i went wrong please?

    Name:  photo.jpg
Views: 734
Size:  341.4 KB

    P.S. sorry i don't know why it's upside down :/
    You can rotate it before you upload
    As for the question, your +11, turned into a -11. . . though you may have made another mistake, I've not checked
    EDIT: I had a look:
    2x^2 \div 2x^2 \not= 0 . . .
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    (Original post by Blueray2)
    Type out the question, I have another method instead of long division, its the bracket method so much easier and quicker, but needs practice
    okay, it's actually a partial fractions questions but as it's a top heavy fraction we got told to use the long division so would mind telling me that way, because that's the way i got taught please?

    the question is:
    Express 2x^2 -x-11/ (2x-3)(x+2) in the form A+ B/2x-3 + C/x+2
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    (Original post by Prista)
    could anyone tell me where i went wrong please?
    Firstly, 2x^2 \div 2x^2 = 1 not 0

    Secondly,  1 \times 2x^2 +x -6 = 2x^2+x-6
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    (Original post by joostan)
    You can rotate it before you upload
    As for the question, your +11, turned into a -11. . . though you may have made another mistake, I've not checked

    Haha it was the right way but then when i uploaded it it rotated
    ahh yeah true, thanks!
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    (Original post by Prista)
    okay, it's actually a partial fractions questions
    In that case you just need that 2x^2 \div 2x^2 = 1 from your long division

    Then do your partial fractions work knowing that A = 1
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    (Original post by TenOfThem)
    Firstly, 2x^2 \div 2x^2 = 1 not 0

    Secondly,  1 \times 2x^2 +x -6 = 2x^2+x-6
    yeah sorry my bad, but then the remainder would be -2x+17 right?
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    (Original post by Prista)
    Haha it was the right way but then when i uploaded it it rotated
    ahh yeah true, thanks!
    For partial fractions like that I'd do it by a bit of algebraic manipulation but then again I'm weird :cool:
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    (Original post by TenOfThem)
    In that case you just need that 2x^2 \div 2x^2 = 1 from your long division

    Then do your partial fractions work knowing that A = 1
    oh okay i see, thank you!
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    (Original post by Prista)
    yeah sorry my bad, but then the remainder would be -2x+17 right?
    But that is irrelevant you only need the long division to get A = 1

    Which, tbh you should just be able to see


    You can split it so that you just use the remainder for your partial fraction work but that is messy and prone to errors
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    (Original post by Blueray2)
    Type out the question, I have another method instead of long division, its the bracket method so much easier and quicker, but needs practice
    I like the bracket method. Its the best (you mean the factorisation starting from 3rd line back to filling up in 2nd ?).
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    (Original post by zaback21)
    I like the bracket method. Its the best (you mean the factorisation starting from 3rd line back to filling up in 2nd ?).
    Only suitable if you are factoring, so not in this question
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    (Original post by zaback21)
    I like the bracket method. Its the best (you mean the factorisation starting from 3rd line back to filling up in 2nd ?).
    Yeah so like you have 2 brackets

    ( fill this is in) (what ever the original thing was)
 
 
 
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