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Q.
Determine the partial fraction decomposition of the given expression $\frac{3 x+17}{x^2+8 x-9}$
Partial Fractions
Solution:
Given: $\frac{3 x+17}{x^2+8 x-9}$
Firstly, let's factor the denominator as much as possible:
$\frac{3 x+17}{(x-1)(x+9)}$
Now the partial fraction decomposition becomes
$\frac{3 x+17}{x^2+8 x-9}=\frac{A}{x-1}+\frac{B}{x+9}$
The LCD for this expression is $(x-1)(x+9)$
$\Rightarrow \frac{3 x+17}{x^2+8 x-9}=\frac{A(x+9)+B(x-1)}{(x-1)(x+9)}$
The denominators are equal on both sides, therefore the numerator becomes
$\Rightarrow 3 x+17=A(x+9)+B(x-1)$
Now we take value of $x$ to determine the constant.
$ x=(-9) \Rightarrow 3(-9)+17=A(-9+9)+B(-9-1) $
$ \Rightarrow 3(-9)+17=-10 B$
$\Rightarrow-27+17=-10 B$
$ \Rightarrow B=1 $
$ x=1 \Rightarrow 3(1)+17=A(1+9)+B(1-1) $
$ \Rightarrow 20=10 A $
$ \Rightarrow A=2$
Required partial fraction decomposition
$=\frac{2}{x-1}+\frac{1}{x+9}$