# Partial fraction

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• August 17th 2011, 07:53 AM
decoy808
Partial fraction
find the values of A,B and C such that

http://quicklatex.com/cache3/ql_4353...003d731_l3.png

i see that it is repeated linear, but the x^2+4 is confusing me. (Worried)
• August 17th 2011, 08:21 AM
FernandoRevilla
Re: Partial fraction
Write

$\frac{2x-3}{(x+1)(x^2+4)}=\frac{A(x^2+4)+(Bx+C)(x+1)}{(x+1) (x^2+4)}$

and identify numerators.
• August 17th 2011, 08:23 AM
TheChaz
Re: Partial fraction
"repeated linear" -?? I've never heard of that. Multiply by the LCD to get
$2x - 3 = A(x^2 + 4) + (Bx + C)(x + 1)$

$2x - 3 = Ax^2 + 4A + Bx^2 + (B + C)x + C$

$2x - 3 = (A + B)x^2 + (B + C)x + (4A + C)$

Now equate coefficients.
• August 17th 2011, 05:29 PM
SammyS
Re: Partial fraction
Quote:

Originally Posted by TheChaz
"repeated linear" -?? I've never heard of that. Multiply by the LCD to get
$2x - 3 = A(x^2 + 4) + (Bx + C)(x + 1)$
...

Equate the numerators in FernandoRevilla's equation, and the result is exactly what you have in the above quote from your post.
• August 17th 2011, 07:46 PM
TheChaz
Re: Partial fraction
True, though I'm not sure I get your point.

Expressing the terms with a common denominator and determining what's left after multiplying by the LCD are procedurally almost identical...
• August 17th 2011, 08:00 PM
SammyS
Re: Partial fraction
Quote:

Originally Posted by TheChaz
True, though I'm not sure I get your point.

Expressing the terms with a common denominator and determining what's left after multiplying by the LCD are procedurally almost identical...

(Blush)Sorry about that Chaz. I didn't look at OP closely enough & thought that somehow you were referring to FernandoRevilla's Post.