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    Try binomial expansion.
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    Try expanding it
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    (Original post by donutellme)
    Try binomial expansion.
    is there no other way of doing it? i mean that would take long to do, no?
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    (Original post by KloppOClock)
    is there no other way of doing it? i mean that would take long to do, no?
    Well you only need to expand things that would result in x^2 s so it shouldn't no
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    (Original post by KloppOClock)
    is there no other way of doing it? i mean that would take long to do, no?
    I assume you know basics of expanding to the degree that it's quite natural. Just discard any terms where the power of x gets greater than 2, and only treat it as if it includes those terms, e.g. 3x(2x(2x^2 + x + 3) goes to 4x^3+2x^2+6x so discard x^3 so you have 3x(2x^2+6x) and so 6x^3+18x^2 and therefore 18x^2
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    (Original post by donutellme)
    I assume you know basics of expanding to the degree that it's quite natural. Just discard any terms where the power of x gets greater than 2
    theres 3 terms tho, i mean i assume i factorise it? but when i did that and then expanded i still got it wrong
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    (Original post by KloppOClock)
    theres 3 terms tho, i mean i assume i factorise it? but when i did that and then expanded i still got it wrong

    I assume you know basics of expanding to the degree that it's quite natural. Just discard any terms where the power of x gets greater than 2, and only treat it as if it includes those terms, e.g. 3x(2x(2x^2 + x + 3) goes to 4x^3+2x^2+6x so discard x^3 so you have 3x(2x^2+6x) and so 6x^3+18x^2 and therefore 18x^2
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    (Original post by donutellme)
    I assume you know basics of expanding to the degree that it's quite natural. Just discard any terms where the power of x gets greater than 2, and only treat it as if it includes those terms, e.g. 3x(2x(2x^2 + x + 3) goes to 4x^3+2x^2+6x so discard x^3 so you have 3x(2x^2+6x) and so 6x^3+18x^2 and therefore 18x^2
    yah u already typed that
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    (Original post by KloppOClock)
    theres 3 terms tho, i mean i assume i factorise it? but when i did that and then expanded i still got it wrong
    Just get the square bracket up to a term of x^2 from binomial expansion or otherwise then expand the two.
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    (Original post by KloppOClock)
    yah u already typed that
    Yeah sorry haha

    Basically, do that method on each of the inside brackets first, then combine them, then do it with the outside bracket, then combine them.
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    Okay I got the right answer but surely im not doing this the correct way?

    So the second part of the bracket can not have any coefficients of x so I just have to expand the first one, so it goes:
    just considering coefficents of upto x^2

    (1+2x+3x^2)(1+2x+3x^2) = 10x^2+4x+1
    (1+4x+10x^2)(1+2x+3x^2) = 21x^2+6x+1
    (1+2x+3x^2) ( 21x^2+6x+1) = (36x^2+8x+1)
    (1+2x+3x^2) (36x^2+8x+1) = 55x^2+20x^2+3x^2 =78x^2

    as the 1 will cancel out with the -1 i only need to work out 78*4 to find the coefficient, which gets the correct answer of 312

    I mean, surely there is a much quicker way of doing this?
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    (Original post by KloppOClock)
    Okay I got the right answer but surely im not doing this the correct way?

    So the second part of the bracket can not have any coefficients of x so I just have to expand the first one, so it goes:
    just considering coefficents of upto x^2

    (1+2x+3x^2)(1+2x+3x^2) = 10x^2+4x+1
    (1+4x+10x^2)(1+2x+3x^2) = 21x^2+6x+1
    (1+2x+3x^2) ( 21x^2+6x+1) = (36x^2+8x+1)
    (1+2x+3x^2) (36x^2+8x+1) = 55x^2+20x^2+3x^2 =78x^2

    as the 1 will cancel out with the -1 i only need to work out 78*4 to find the coefficient, which gets the correct answer of 312

    I mean, surely there is a much quicker way of doing this?
    That seems about right. You can eliminate even more possibilities by sight, but that comes by experience.
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    (Original post by donutellme)
    That seems about right. You can eliminate even more possibilities by sight, but that comes by experience.
    how am i meant to do that in under 3 minutes without a calculator, it took me like 5 just to write out the answer, smh
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    (Original post by KloppOClock)
    how am i meant to do that in under 3 minutes without a calculator, it took me like 5 just to write out the answer, smh
    I'll try it tomorrow for myself and time it and show you my working if you like? Practice will make you better.
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    (Original post by KloppOClock)
    Okay I got the right answer but surely im not doing this the correct way?

    So the second part of the bracket can not have any coefficients of x so I just have to expand the first one, so it goes:
    just considering coefficents of upto x^2

    (1+2x+3x^2)(1+2x+3x^2) = 10x^2+4x+1
    (1+4x+10x^2)(1+2x+3x^2) = 21x^2+6x+1
    (1+2x+3x^2) ( 21x^2+6x+1) = (36x^2+8x+1)
    (1+2x+3x^2) (36x^2+8x+1) = 55x^2+20x^2+3x^2 =78x^2

    as the 1 will cancel out with the -1 i only need to work out 78*4 to find the coefficient, which gets the correct answer of 312

    I mean, surely there is a much quicker way of doing this?
    You didn't use the binomial expansion.

    (1+2x+3x^2)^6=(1+[2x+3x^2])^6

    =1 + 6[2x+3x^2]+\frac{6\times 5}{2}[2x+3x^2]^2+...

    and that's as far as you need to go.

    =1 + 18x^2+15\times 4x^2+...

    Edit: Ignoring terms that aren't relevant.
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    The answer is 312 - you are correct. But, the faster method would be to follow Ghostwalker ^
 
 
 
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