Partial Fractions Question

**Paymemoney** · March 26th 2010, 01:27 AM

Hi
Need help on the following questions:
1) Resolve into partial fractions: $\frac{x^2-x+1}{(4x^2-1)(x-2)}$

This is what i have done:
$\frac{x^2-x+1}{(4x^2-1)(x-2)}$ => $\frac{x^2-x+1}{(x-2)(2x-1)(2x+1)}$

$\frac{A}{(x-2)}+\frac{B}{(2x-1)}+\frac{C}{(2x+1)}$

$A(2x-1)(2x+1)+B(x-2)(2x+1)+C(x-2)(2x-1)=x^2-x+1$

$A(4x^2-1)+B(2x^2-3x-2)+C(2x^2-5x+2)=x^2-x+1$

Let x = 2

$15A=1$

$A=\frac{1}{15}$

$Let x = \frac{1}{2}$

$\frac{-6}{2}B=1$

$B=\frac{-2}{6}$

Let x= $\frac{-1}{2}$

$\frac{10}{2}C=1$

$C=\frac{1}{5}$

$\frac{\frac{1}{15}}{(x-2)}+\frac{\frac{-2}{6}}{(2x-1)}+\frac{\frac{1}{5}}{(2x+1)}$

book's answer says it $\frac{\frac{1}{5}}{(x-2)}-\frac{\frac{1}{4}}{(2x-1)}+\frac{\frac{7}{20}}{(2x+1)}$

someone tell me where i have gone wrong.

P.S

**zhangyi05** · March 26th 2010, 03:17 AM

let A=2
then 15=3A not 15=A

**HallsofIvy** · March 26th 2010, 04:06 AM

Zhangyi05 means "Let x= 2".

**Paymemoney** · March 26th 2010, 04:24 AM

thanks i understand now.

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