# I'm having a problem with the following partial fraction

• Dec 9th 2009, 01:07 PM
swatpup32
I'm having a problem with the following partial fraction
http://img707.imageshack.us/img707/1045/71577508.jpg
Why is it Ax(x+3) + B(x+3) + Cx^2
instead of Ax^2(x+3) + Bx(x+3) + Cx^3
• Dec 9th 2009, 01:11 PM
Raoh
Quote:

Originally Posted by swatpup32
http://img707.imageshack.us/img707/1045/71577508.jpg
Why is it Ax(x+3) + B(x+3) + Cx^2
instead of Ax^2(x+3) + Bx(x+3) + Cx^3

hello(Hi)
$\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+3}=\frac{Ax(x +3)+B(x+3)+Cx^2}{x^2(x+3)}$
• Dec 9th 2009, 01:23 PM
swatpup32
I'm sorry, but I still don't understand.
• Dec 9th 2009, 01:25 PM
Raoh
Quote:

Originally Posted by swatpup32
I'm sorry, but I still don't understand.

do you know how to make a single fraction of this expression, $\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+3}$ ?
• Dec 9th 2009, 01:31 PM
swatpup32
okay, I wasn't thinking about it as a fraction. I was only considering the numerator. An, x canceled out from the numerator and denominator.
• Dec 9th 2009, 01:35 PM
Raoh
Quote:

Originally Posted by Raoh
do you know how to make a single fraction of this expression, $\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+3}$ ?

you must make $\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+3}$ a single fraction with the denominator $x^2(x+3)$.
• Dec 9th 2009, 01:40 PM
Raoh
$\frac{x^2+8x-3}{x^2(x+3)}$ $=\frac{Ax(x+3)+B(x+3)+Cx^2}{x^2(x+3)}$
now get rid of the same denominators and you'll have,
$x^2+8x-3=Ax(x+3)+B(x+3)+Cx^2$