Binomial Expansion

Ok so part b (June 2006, Q2) wants me to find the coefficient of the x^2 term of

\frac{(1+2x)^{2}}{(1-3x)^{2}}

. The bottom is equal to

27x^{2}+6x+1

when expanded by binomial in part a...

So far I have

\frac{4x^{2}+4x+1}{27x^{2}+6x+1}

What should I do now to work out the coefficient?

(edited 13 years ago)

Reply 1

Daniel Freedman

You don't want to expand the denominator like that. For these questions, write it as

(1+2x)^2 (1-3x)^{-2}

. Expand the second bracket using the binomial theorem, and the first normally.

Reply 2

Stephhcharlene

Oh i made silly mistake.. yeah I forgot that the bottom is already to -2 not 2... therefore already in your form as such.. :P I just didnt engage my brain properly! lol

Reply 3

In One Ear

Original post by Stephhcharlene

The bottom is equal to

27x^{2}+6x+1

when expanded by binomial in part a..

What should I do now to work out the coefficient?

For a start im afraid that the bottom is equal to

9x^2-6x+1

Reply 4

Stephhcharlene

Original post by In One Ear

For a start im afraid that the bottom is equal to

9x^2-6x+1

I said I wrote it wrong. to the power of -2 is 27x^2 :smile:

I was right I just jumbled my working out wrong and forgot that the question said positive 2. :P

Reply 5

In One Ear

Oh sorry, i had left my response without posting for a while so i didnt see the previous responses. My mistake :redface:

Quick Reply

Latest

Articles for you

How to revise for GCSE Psychology exams: AQA explains what to do

Postgraduate Doctoral Loan eligibility

Nationality and residency

Postgraduate Master’s Loan

Binomial Expansion

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you