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Core 4 help!!! binomial expansion with fractions.

just wondering if anyone could explain this to me.

Given that 2x^2-3 / (3-2x) (1-x)^2 can be written in the form A / (3-2x) + B / (1-x) + C / (1-x)^2
find the values of A, B, and C.

Hence find the binomial expansion of 2x^2-3 / (3-2x) (1-x)^2 up to x^3.
??

anyone???

First bit you need to split into partial fractions, then once you've done this, you can re-write the fractions as a(3-2x)^-1 and then do the expansion as normal.

Once you've expanded them all, you then add all the terms together to get your final result

Reply 3

Jfromage

but how would i do that?
i get confused when there are 3 different fractions i can do it for two .
thanks

Reply 4

zahre

Original post by Jfromage

2x^2-3/(3-2x)(1-x)^2 = A/(3-2X) + B/(1-X) + C/(1-X)^2

2x^2-3 = A/(3-2X) + B/(1-X) + C/(1-X)^2

When x = 1
-1=1C
so C = -1

When x = 3/2
3/2 = 1/4A
so A = 6

Then you put in x = 0 into it which I think is where you got stuck.
When x=0

-3 = A + 3B + 3C
then substitute in your values for A and C to find B.

So -6 = 3B
B = -2

So in the end you get 6/(3-2x) - 2/(1-x) - 1/(1-x)^2. Which I think should be correct. :biggrin: