Question

Determine the partial fraction decomposition of the given expression $\frac{3x+17}{x^2+8x-9}$

• A. $\frac{2}{x-1}+\frac{1}{x+9}$
• B. $\frac{1}{x-1}+\frac{2}{x+9}$
• C. $\frac{2}{x+1}+\frac{1}{x-9}$
• D. none

Answer 1

Answer: (A)

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