# Partial Fractions Question

• Mar 26th 2010, 01:27 AM
Paymemoney
Partial Fractions Question
Hi
Need help on the following questions:
1) Resolve into partial fractions: $\displaystyle \frac{x^2-x+1}{(4x^2-1)(x-2)}$

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

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

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

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

Let x = 2

$\displaystyle 15A=1$

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

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

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

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

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

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

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

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

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

someone tell me where i have gone wrong.

P.S
• Mar 26th 2010, 03:17 AM
zhangyi05
let A=2
then 15=3A not 15=A
(Wondering)
• Mar 26th 2010, 04:06 AM
HallsofIvy
Zhangyi05 means "Let x= 2".
• Mar 26th 2010, 04:24 AM
Paymemoney
thanks i understand now.