I need help with these...

Partial Fraction Decomposition:

1/4x^2-9

and

-2x+15/x^2-x-12

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- Oct 28th 2009, 04:43 AMRenSullyPartial Fraction Decompostion
I need help with these...

Partial Fraction Decomposition:

1/4x^2-9

and

-2x+15/x^2-x-12 - Oct 28th 2009, 05:48 AMmasters
Hi RenSully,

Here's the second one:

If all the factors are of the form (*x*-*c*), then each of the partial fractions will be of the form $\displaystyle \frac{A}{x-c}$.

$\displaystyle \frac{-2x+15}{x^2-x-12}=\frac{-2x+15}{(x-4)(x+3)}\Rightarrow\frac{A}{x-4}+\frac{B}{x+3}$

First, multiply the original proper fraction by (*x*-*c*) and evaluate the result at*x*=*c*to get the value of the constant.

$\displaystyle \frac{-2x+15}{(x-4)(x+3)}(x-4)=\frac{-2x+15}{x+3}\Rightarrow A=\frac{-2(4)+15}{4+3}=1$

$\displaystyle \frac{-2x+15}{(x-4)(x+3)}(x+3)=\frac{-2x+15}{x-4}\Rightarrow B=\frac{-2(-3)+15}{-3-4}=-3$

Thus:

$\displaystyle \frac{-2x+15}{x^2-x-12}=\frac{1}{x-4}+\frac{-3}{x+3}$