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    (Original post by Hamzah249)
    I don't know why you are getting all worked up on those trying to help you, we did look at the picture maybe you need to read what we said?
    You're not exactly helping. I've said I'm stuck
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    (Original post by Hamzah249)
    I don't know why you are getting all worked up on those trying to help you, we did look at the picture maybe you need to read what we said?
    lol exactly. As we said one can't compare them so simply as it requires some expanding since B isn't the only thing with coefficient x
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    (Original post by Student1256)
    lol exactly. As we said one can't compare them so simply as it requires some expanding since B isn't the only thing with coefficient x
    You didn't say any of that so don't lie?
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    (Original post by Jane122)
    You're not exactly helping. I've said I'm stuck
    Listen. Take the whole equation and multiply A and B into the brackets. Then compare coefficients of x with x and coefficients of x^2 with x^2.
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    (Original post by Student1256)
    Listen. Take the whole equation and multiply A and B into the brackets. Then compare coefficients of x with x and coefficients of x^2 with x^2.
    Would I follow this method all the time with two brackets ? And if there were more than two brackets would I multiply all of them out like this ?
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    Troll?
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    (Original post by Jane122)
    You didn't say any of that so don't lie?
    ...
    (Original post by Student1256)
    You're comparing the coefficients wrong. You can't simply compare A as the coefficient of the x to the power 2
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    (Original post by Jane122)
    Would I follow this method all the time with two brackets ? And if there were more than two brackets would I multiply all of them out like this ?
    yes try it and you'll get the correct answer. When you expand you'll find that A and B both have the coefficient x so you'll need to compare them both not just B. You can solve it simultaneously as A = 3 which you're correct in I believe.
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    (Original post by Student1256)
    ...
    Exactly, I even gave him a tip if comparing coefficients is too hard, simply take x=1 or 2 etc
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    (Original post by Jane122)
    ..
    LHS is (Ax^2+4Ax+4A)+(Bx+2B)+C=(A)x^2+(  4A+B)x+(4A+2B+C)

    Comparing coefficients with RHS gives you:
    x^2 : A=3
    x : 4A+B=12
    c : 4A+2B+C=8

    which I am sure you can solve very easily. Doing it this way is the proper way. Noticing substitutions like x=-2 is merely a shortcut which doesnt always spit out all the values you need.
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    (Original post by Jane122)
    You're not exactly helping. I've said I'm stuck
    Please expand A(x + 2)(x + 2)
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    (Original post by Hamzah249)
    Exactly, I even gave him a tip if comparing coefficients is too hard, simply take x=1 or 2 etc
    Im a female. So shut up.
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    (Original post by RDKGames)
    LHS is (Ax^2+4Ax+4A)+(Bx+2B)+C=(A)x^2+(  4A+B)x+(4A+2B+C)

    Comparing coefficients with RHS gives you:
    x^2 : A=3
    x : 4A+B=12
    c : 4A+2B+C=8

    which I am sure you can solve very easily. Doing it this way is the proper way. Noticing substitutions like x=-2 is merely a shortcut which doesnt always spit out all the values you need.
    Thank you so much !!!! Xxx
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    (Original post by RDKGames)
    LHS is (Ax^2+4Ax+4A)+(Bx+2B)+C=(A)x^2+(  4A+B)x+(4A+2B+C)

    Comparing coefficients with RHS gives you:
    .
    Please don't post full solutions - the OP would have got there
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    Just work out The right Hand side in order And equate The terms on The LEFT HAND side. To DECODE The fraction you had at The start.
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    Name:  IMG_4895.jpg
Views: 11
Size:  450.9 KBAttachment 612454612456



    Guys help!!! How do I do part b?

    Muttley79 RDKGames Hamzah249




    Attachment 612458 Would I only do partial fractions on 0/x+3
    Attached Images
     
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    (Original post by RDKGames)
    LHS is (Ax^2+4Ax+4A)+(Bx+2B)+C=(A)x^2+(  4A+B)x+(4A+2B+C)

    Comparing coefficients with RHS gives you:
    x^2 : A=3
    x : 4A+B=12
    c : 4A+2B+C=8

    which I am sure you can solve very easily. Doing it this way is the proper way. Noticing substitutions like x=-2 is merely a shortcut which doesnt always spit out all the values you need.
    Sir how would you make partial fractions of this (x^2)/(x+1)? Thank you
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    (Original post by Jane122)

    Guys help!!! How do I do part b?

    Muttley79 RDKGames

    Would I only do partial fractions on 0/x+3
    0/x+3 ?

    Just factorise f(x) completely then express \frac{10}{f(x)}=\frac{10}{(x-a)(x-b)(x-c)} then in partial fraction form where a,b,c are the roots
    before finding the values of A, B and C.
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    (Original post by RDKGames)
    0/x+3 ?

    Just factorise f(x) completely then express \frac{10}{f(x)}=\frac{10}{(x-a)(x-b)(x-c)} then in partial fraction form where a,b,c are the roots
    before finding the values of A, B and C.
    Can you look at my third attachment please
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    (Original post by Student1256)
    Sir how would you make partial fractions of this (x^2)/(x+1)? Thank you
    Ax+B+\frac{C}{x+1}
 
 
 
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