# partial fraction

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#1
how would you solve 3x/(2-x)(4+x^2)??? i tried looking at the mark scheme and it made no sense!!
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1 month ago
#2
Compare coeffefciients
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1 month ago
#3
(Original post by nomen552)
how would you solve 3x/(2-x)(4+x^2)??? i tried looking at the mark scheme and it made no sense!!
What do you mean by "solve"? Do you mean writing it in terms of partial fractions?

Where exactly are you stuck / what techniques do you know for decomposing something like this into partial fractions?
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1 month ago
#4
What I would do is:a/(2-x)+b(4+x^2) then you do a(4+x^2)+b(2-x) all /(2-x)(4+x^2)=3x/(2-x)(4+x^2).Then you can work it out from there. I would suggest using mathswatch.
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1 month ago
#5
4-x^2 is an irreducible quadratic factor

So the form of your solution should be
A/(2-x) + (Bx+C) /(4x^2)

The you can try comparing coefficients.

If you post what the solution says then someone could likely explain that to you.
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