Partial Fractions Question

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January 4th 2010, 04:22 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
January 4th 2010, 04:30 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.
January 4th 2010, 04:36 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 $\frac{x}{(x+1)(x-2)}$
January 4th 2010, 04:45 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 $\frac{x}{x^2-x-2}$
January 4th 2010, 04:49 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.

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