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    C3 Heinemann Modular, page 8 [Example 12]

    So i can work out C and D, but i can't figure out how 1 = A and then B = 4

    They're saying to compare co-efficients but i don't understand. help please.
    Can't move on till i work this out..
    I haven't written out the question because i don't know how to use Latex or whatever it's called, sorry.

    Rei.
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    Well given hardly any of us are going to have the book to hand I suggest you do your best to reproduce the question.
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    (Original post by Rei)
    C3 Heinemann Modular, page 8 [Example 12]

    So i can work out C and D, but i can't figure out how 1 = A and then B = 4

    They're saying to compare co-efficients but i don't understand. help please.
    Can't move on till i work this out..
    I haven't written out the question because i don't know how to use Latex or whatever it's called, sorry.

    Rei.
    There's a link above your post called "How to use LaTeX". Have a look at it! The equation can't be that bad that it'll take you ages to figure out how to post it in Latex.
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    take a picture or scan it
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    I don't have the question.

    But, if you're dividing something by one of its factors, the remainder must equal 0. So if you're remainder is something like ax^2 + bx + c, then what you do is say ax^2 + bx + c = 0x^2 + bx + c, and then you equate like with like. So  ax^2 = 0x^2 \implies a = 0 and so on.

    Do you understand?

    And if you know that x - 3 is a factor and hence f(3) = 0 or whatever then it's the same dealy. x terms = x terms, x^2 terms = x^2 terms and such.

    I dunno what the question is asking or anything but yeah.
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    http://tinypic.com/view.php?pic=in8zv6&s=3
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    When you open up the bracket, you have to look at the matching variables:

    x^3 = Ax^3
    So when we look at the co-efficient (The numbers in front of x^3), we get 1 = A

    Similarly,
    x^2 = -3Ax^2 + Bx^2
    So when we look at the co-efficient (The numbers in front of x^2), we get 1 = -3A + B but since we know what A is then we can substitute its value to get 

1 = -3A + B \implies 1 = -3(1) + B \implies B = 4.
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    Sorry, i didn't make this clear enough.. i understand what a co-efficient is. I just don't understand how the were able to equate x^3 to Ax^3 when in the question it's Ax^2

    Same with B and the x^2
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    (Original post by Rei)
    Sorry, i didn't make this clear enough.. i understand what a co-efficient is. I just don't understand how the were able to equate x^3 to Ax^3 when in the question it's Ax^2.

    Same with B and the x^2.
    It's not Ax^2. It's (Ax^2 + stuff)(x + stuff), which multiplies out to give you Ax^3. (If I read the question correctly.)
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    (Original post by generalebriety)
    It's not Ax^2. It's (Ax^2 + stuff)(x + stuff), which multiplies out to give you Ax^3. (If I read the question correctly.)
    So, you multiply by x for A, then simply compare those co-efficients. What about for B, you multiply by -3? if so, how comes.. ?
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    (Original post by Rei)
    So, you multiply by x for A, then simply compare those co-efficients. What about for B, you multiply by -3? if so, how comes.. ?
    Ho hum.

    Multiply out the brackets on the RHS, it'll make the whole thing a lot clearer to you. All they're doing is multiplying the brackets out systematically in their heads.
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    (Original post by generalebriety)
    Ho hum.

    Multiply out the brackets on the RHS, it'll make the whole thing a lot clearer to you. All they're doing is multiplying the brackets out systematically in their heads.
    LOL, I just looked at it again.. figured out what the hell was wrong with me.. Ahhhh for some stupid reason i kept working the expansion out mentally and coming to -2Ax^3 when i multiplied by -3. Thanks for that.
 
 
 
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