# Partial Fractions Question

• Jan 4th 2010, 05:22 PM
Paymemoney
Partial Fractions Question
Hi
Need help to express the following in partial fractions:

$\frac{x^2-2}{(x+1)(x-2)}$

P.S
• Jan 4th 2010, 05:30 PM
pickslides
What do you need to do with this expression?

The order of the numerator is the same as the denominator, you may not need to make partial fractions.

If you need to consider

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

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

choose values for x to solve for A and B.
• Jan 4th 2010, 05:36 PM
skeeter
Quote:

Originally Posted by Paymemoney
Hi
Need help to express the following in partial fractions:

$\frac{x^2-2}{(x+1)(x-2)}$

P.S

$\frac{x^2-2}{(x+1)(x-2)} =$

$\frac{x^2-2}{x^2-x-2} =$

$\frac{x^2-x-2+x}{x^2-x-2} =$

$\frac{x^2-x-2}{x^2-x-2} + \frac{x}{x^2-x-2} =$

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

now use the method of partial fractions on $\frac{x}{(x+1)(x-2)}$
• Jan 4th 2010, 05:45 PM
Paymemoney
Quote:

Originally Posted by skeeter

$\frac{x^2-x-2}{x^2-x-2} + \frac{x}{x^2-x-2} =$

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

Can you explain to me why did you plus $\frac{x}{x^2-x-2}$
• Jan 4th 2010, 05:49 PM
skeeter
Quote:

Originally Posted by Paymemoney
Can you explain to me why did you plus $\frac{x}{x^2-x-2}$

and then how did you get $1 + \frac{x}{(x+1)(x-2)}$

note that the numerator and denominator of your original fraction have the same degree.

you can get the same result by dividing $x^2-2$ by $(x+1)(x-2)$ using long division.