# Expressing in partial fractions

• Jul 7th 2007, 11:11 AM
erika
Expressing in partial fractions
$\displaystyle (9x^2) / [(x -1)^2 * (x + 2)]$

It's part of a set of partial fractions I have to do.. I can't seem to solve this?
• Jul 7th 2007, 11:13 AM
Jhevon
Quote:

Originally Posted by erika
$\displaystyle (9x^2) / [(x -1)^2 * (x + 2)]$

It's part of a set of partial fractions I have to do.. I can't seem to solve this?

$\displaystyle \frac {9x^2}{(x - 1)^2 (x + 2)} = \frac {A}{x - 1} + \frac {B}{(x - 1)^2} + \frac {C}{x + 2}$

Can you take it from here?
• Jul 7th 2007, 11:15 AM
erika
Meh I did that and then tried substituting values for $\displaystyle x$ but it didn't seem to work :confused:
• Jul 7th 2007, 11:18 AM
Jhevon
Quote:

Originally Posted by erika
Meh I did that and then tried substituting values for $\displaystyle x$ but it didn't seem to work :confused:

Substitute values of x? do you know the method of partial fractions?

multiply through by the denominator of the left side, we get:

$\displaystyle 9x^2 = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)^2$

Can you take it from here? You can multiply out and equate coefficients, or you can choose convenient values of x that wipes out two variables at a time
• Jul 7th 2007, 11:35 AM
erika
http://www.mathhelpforum.com/math-he...5d4e11b0-1.gif

sub x: 1
9 = 3B
3 = B

sub x: -1
36 = 9C
4 = C

By comparing coefficient of x^2:
9 = A + C
A = 5

Is that correct?
• Jul 7th 2007, 11:41 AM
Jhevon
Quote:

Originally Posted by erika
http://www.mathhelpforum.com/math-he...5d4e11b0-1.gif

sub x: 1
9 = 3B
3 = B

sub x: -1
36 = 9C
4 = C

By comparing coefficient of x^2:
9 = A + C
A = 5

Is that correct?

Yes! Good Job! :)
• Jul 7th 2007, 11:58 AM
erika
Thank you!
One last question...

$\displaystyle \frac {17x^2+23x+12}{(3x+4)(x^2+4)} = \frac{A}{3x+4} + \frac{B}{x^2+4}$

By cover-up rule, I got
$\displaystyle A = \frac {17(-4/3)^2+23(-4/3)+12}{(-4/3)^2+4} = 2$

But since$\displaystyle (x^2 +4)$ isn't linear, does it mean I follow the steps you mentioned earlier (multiply through by denominator, etc)?
• Jul 7th 2007, 12:01 PM
Jhevon
Quote:

Originally Posted by erika
Thank you!
One last question...

$\displaystyle \frac {17x^2+23x+12}{(3x+4)(x^2+4)} = \frac{A}{3x+4} + \frac{B}{x^2+4}$

By cover-up rule, I got
$\displaystyle A = \frac {17(-4/3)^2+23(-4/3)+12}{(-4/3)^2+4} = 2$

what is this "cover-up" rule you are referring to?

Quote:

But since$\displaystyle (x^2 +4)$ isn't linear, does it mean I follow the steps you mentioned earlier (multiply through by denominator, etc)?
yes, but you have to change the numerator. The numerator must always be one degree less than the denominator. so for the part where the denominator is $\displaystyle x^2 + 4$ you must write:

$\displaystyle \frac {Bx + C}{x^2 + 4}$