# Thread: Quadratic Expansion using Binomial Theorem

1. ## Quadratic Expansion using Binomial Theorem

Hi, I am stuck with a question about Binomial Expansion using Binomial Theorem, here is the question:

Find, in ascending powers of x, the first 4 terms in the expansion of (1 - 2x - 4x^2)^11

Here is how I work out the question:
[1 - 2x(1 + 2x)]^11
= 1 + (11C1)[-2x(1+2x)] + (11C2)[-2x(1+2x)]^2 + (11C3)[-2x(1+2x)]^3 + .........
= 1 + 11[-2x(1+2x)] + 55[-2x(1+2x)]^2 + 165[-2x(1+2x)]^3 + ............
= 1 - 22x - 44x^2 + 55[(2x)^2 - 2(2x)(4x^2) + (4x^2)^2] + 165(-2x - 4x)^3 + ....
= 1 - 22x - 44x^2 + 220x^2 - 880x^3 +880x^4
+ 165[(-2x)^3 + 3(-2x)^2(-4x^2) + 3(-2x)(-4x^2)^2 + (-4x^2)^2
= 1 - 22x - 44x^2 + 220x^2 - 880x^3 +880x^4
+ 165(-8x^3 + 48x^4 + 96x^5 - 64x^6)
= 1 - 22x + 176x^2 - 880x^3 + 880x^4 - 1320x^3 + ...............
= 1 - 22x + 176x^2 - 2200x^3

The bold part is the one that is wrong, I am not sure how I do it wrongly, can anyone guide me? thanks

the solution is 1 - 22x + 176x^2 - 440x^3 + .......

2. ## Re: Quadratic Expansion using Binomial Theorem

Welcome to MHF, danny2012!

Originally Posted by danny2012
= 1 + 11[-2x(1+2x)] + 55[-2x(1+2x)]^2 + 165[-2x(1+2x)]^3 + ............
= 1 - 22x - 44x^2 + 55[(2x)^2 - 2(2x)(4x^2) + (4x^2)^2] + 165(-2x - 4x)^3 + ....[tex]
You kept the minus sign in this step when you shouldn't have.

$[-2x(1+2x)]^2 = (2x)^2(1+4x+4x^2)$

3. ## Re: Quadratic Expansion using Binomial Theorem

Hi ILikeSerena, thanks for your warm welcome

You kept the minus sign in this step when you shouldn't have.
[-2x(1+2x)]^2 = (2x)^2(1+4x+4x^2)
I see, I forgot to include the bracket in -2x^2 which should be (-2x)^2 and the answer is 4x^2 lol thanks for the point out ILikeSerena

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