Binomial Expansion Teaches how to choose the RIGHT partner (:

I always like to share with my students what each Math topic has to do with their everyday life, particularly their future.

Yesterday, I did Binomial Expansion. It's about finding the RIGHT partner (: It's about mastering the skills of finding the correct match based on a set of factors.

Just like this question, you see that there are 2 brackets. 1st bracket is good for now. 2nd bracket needs us to do some expansion work by using the FORMULAE (It's provided during GCE O Level Exams but some schools don't seem to provide it for their mid year, strange)
Oh yar, there is NO need to remember the Binomial Expansion Formulae!
Then one of the questions that often pop up is " When do we stop our expansion for [TEX](1+\frac {x}{3})^{12}[/TEX]?"

To answer this question, it depends on what type of partners you are looking for.


We are interested in the coefficients of x & x^2 so from the first bracket :

To have x,

  • 6 will pair up with x from [TEX](1+\frac {x}{3})^{12}[/TEX]
  • 2x will pair up with the constant from [TEX](1+\frac {x}{3})^{12}[/TEX]

Similarly, to have x^2,

  • 6 will pair up with x^2 from [TEX](1+\frac {x}{3})^{12}[/TEX]
  • 2x will pair up with the x,from [TEX](1+\frac {x}{3})^{12}[/TEX]
  • -3x^2 will pair up with the constant from [TEX](1+\frac {x}{3})^{12}[/TEX]

Now, do you know where you stop the expansion of [TEX](1+\frac {x}{3})^{12}[/TEX]?

Stop at the x^2 term :) aka the 3rd term

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Hi, I'm Ai Ling Ong. I enjoy coaching students who have challenges with understanding and scoring in 'O' Level A-Maths and E-Maths. I develop Math strategies, sometimes ridiculous ideas to help students in understanding abstract concepts the fast and memorable way. I write this blog to share with you the stuff I teach in my class, the common mistakes my students made, the 'way' to think, analyze... If you have found this blog post useful, please share it with your friends. I will really appreciate it! :)

2 Responses to Binomial Expansion Teaches how to choose the RIGHT partner (:

  1. How does the 6 pair up with the x^2 in (1+x/3)^12? there is no x^2, only an x, surely if you times them together you won't get a coefficient for the x^2 value, but for the x.... Can you please explain :/


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