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So confused with partial fractions 3 linear factor [one repeated] ???


But isn't it suppose to be.. :s-smilie:
A(x+1)(x+1)2 + B(2x+1)(x+1)2 +C(2x+1)(x+1)
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Reply 1
Avatar for Alexion
Alexion
17
No. Look at how that step works.

You're multiplying every term in the top line by (2x + 1)(x + 1)2.

Look at what that gives you.
Reply 2
Avatar for Adorable98
Adorable98
OP
15
Original post by Alexion
No. Look at how that step works.

You're multiplying every term in the top line by (2x + 1)(x + 1)2.

Look at what that gives you.


Okay, I get it now! Thanks!!
Reply 3
Avatar for nerak99
nerak99
13
In other words, what you are doing is rather like adding together a quarter and a sixth and using 24 as the common denominator when the "lowest" common denominator would be 12.

You technique will work but will have a cubic terms in the coefficient matching whose coefficient wil;l then be zero. i.e. you will end up doing more work than you have to.
Reply 4
Avatar for Adorable98
Adorable98
OP
15
Original post by nerak99
In other words, what you are doing is rather like adding together a quarter and a sixth and using 24 as the common denominator when the "lowest" common denominator would be 12.

You technique will work but will have a cubic terms in the coefficient matching whose coefficient wil;l then be zero. i.e. you will end up doing more work than you have to.

I see, thank you!!

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