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# maffs question

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1. Try binomial expansion.
2. Try expanding it
3. (Original post by donutellme)
Try binomial expansion.
is there no other way of doing it? i mean that would take long to do, no?
4. (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
5. (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
6. (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
7. (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
8. (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
9. (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 from binomial expansion or otherwise then expand the two.
10. (Original post by KloppOClock)
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.
11. 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?
12. (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.
13. (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
14. (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.
15. (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.

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

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

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