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Nylex
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#21
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#21
(Original post by TheWolf)
ahh cool i think you meant quotient for the latter - but yea thanks!
Yeah, I did. Edited it now.
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kikzen
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#22
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#22
(Original post by mik1a)
Express as partial fractions:

(x^2 + 1)/(x^2 - 1)

I dunno how to do long division with a quadratic denominator, and the fraction is improper as it is now. Please help.
with these ones find a way to make the change the bottom to the top

that sounds a bit confusing

ok
(x^2 - 1) + something = (x^2 + 1)

this one is quite easy, its 2

so the fraction is
[(x^2 - 1) + 2] / (x^2 - 1)

= 1 + 2/(x+1)(x-1)

which you can do as normal
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thenarbisbanned
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#23
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#23
(Original post by kikzen)
with these ones find a way to make the change the bottom to the top

that sounds a bit confusing

ok
(x^2 - 1) + something = (x^2 + 1)

this one is quite easy, its 2

so the fraction is
[(x^2 - 1) + 2] / (x^2 - 1)

= 1 + 2/(x+1)(x-1)

which you can do as normal
I dont know if youve noticed but weve already done/explained this?
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kikzen
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#24
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#24
(Original post by thenarbisbanned)
I dont know if youve noticed but weve already done/explained this?
whoops.

i thought about reading through to check if it had been suggested but then i thought.. hmm well if someones not posted it then it'd be good but if they have then at least i tried to do my bit ;p

(plus reading thru sounded like a bit too much work)
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TheWolf
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#25
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#25
express as partial fractions:

3/ (x³) + 1
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Euler
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#26
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#26
(Original post by TheWolf)
express as partial fractions:

3/ (x³) + 1

do u mean?

3/(x³ +1)

OR

(3/x³) + 1

??
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TheWolf
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#27
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#27
(Original post by integral_neo)
do u mean?

3/(x³ +1)

OR

(3/x³) + 1

??
ahh dont worry i know how to do it now, just have to factorize the bottom
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JamesF
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#28
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#28
x^3 + 1 = (x+1)(x^2 - x + 1)
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JamesF
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#29
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#29
(Original post by TheWolf)
ahh dont worry i know how to do it now, just have to factorize the bottom
Whoops didnt read that.
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