$\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?

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- Jul 7th 2007, 11:11 AMerikaExpressing 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 AMJhevon
- Jul 7th 2007, 11:15 AMerika
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 AMJhevon
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 AMerika
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 AMJhevon
- Jul 7th 2007, 11:58 AMerika
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 PMJhevon
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)?

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