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# Partial Fractions watch

1. Hi

I am going through a chronic mental block =/
well its just questions where the denominator has something to the degree where i stuck at.
for example:
Find the binomial expansion of 3+2x^2 / (2x+1) (x-3)^2
now i know first i need express them as partial fractions.
3+2x^2 similar to A/(2x+1) + B(x-3)^2
now when solve the numerator/ denominator i get
3+2x^2 similar to A(x-3)^2 + B(2x+1) / (2x+1)(x-3)^2
now i know to get the co-efficient values i cant let (2x+1)= 0 to get the value of x=-1/2 for A , but when it comes to go for the value of B i am lost =/.
2. (Original post by ibysaiyan)
3+2x^2 similar to A/(2x+1) + B(x-3)^2
The word/phrase/meaning isn't "similar to", it's "identical to".

When you have powers, you need to do this:

e.g. f(x)/(x+6)^4 = A/(x+6) + B/(x+6)^2 + C/(x+6)^3 + D/(x+6)^4

So lots of partial fractions up to and including the highest power.
3. (Original post by Swayum)
The word/phrase/meaning isn't "similar to", it's "identical to".

When you have powers, you need to do this:

e.g. f(x)/(x+6)^4 = A/(x+6) + B/(x+6)^2 + C/(x+6)^3 + D/(x+6)^4

So lots of partial fractions up to and including the highest power.
Thanks for replying, yea i know that but i have no idea on questions where say for example: A/x-3 + B/x + C/(x-4) + D/(x-4)^2

now i know value of x in A = 3, B= 0 , C= 4 , D= no idea <-- do i compare co-efficients or something else?
4. Yes, compare coefficients of LHS and RHS - you sometimes have to form simultaneous equations even to solve them.
5. (Original post by Swayum)
Yes, compare coefficients of LHS and RHS - you sometimes have to form simultaneous equations even to solve them.
Sorry to bother you, but could you enlighten me a bit more on comparing co-efficients. I tried to look for a particular example in the book but it doesn't have one , all it said was to compare them , haven't got a clue on how to do that.
Thanks again for the reply. =)
6. Ah nevermind i found it =), thanks though.

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