Chapter 7.3, Problem 80E

Solution

To find: The partial fraction decomposition of $\frac{3x^2+2x}{x^2-4}$ and compare the graph.

The required partial fraction decomposition is $2+\frac{3}{x-1}-\frac{2}{x+1}$ .

Given information:

The given rational function is $\frac{3x^2+2x}{x^2-4}$ .

Explanation:

Consider the given rational function.

By using the division:

  

So,

  $\begin{array}{c}\frac{2x+12}{x^2-4}=\frac{A}{x-2}+\frac{B}{x+2}\\ 2x+12=A\left(x+2\right)+B\left(x-2\right)\end{array}$

Now compare the coefficient and constants.

  $\begin{array}{c}A+B=2\\ 2A-2B=12\end{array}$

Write the system as matrix equation $C=\left[\begin{array}{cc}1& 1\\ 2& -2\end{array}\right],X=\left[\begin{array}{c}A\\ B\end{array}\right]\text{ and }D=\left[\begin{array}{c}2\\ 12\end{array}\right]$

Now solve the equation $CX=D$ .

  $\begin{array}{c}\left[\begin{array}{cc}1& 1\\ 2& -2\end{array}\right]\left[\begin{array}{c}A\\ B\end{array}\right]=\left[\begin{array}{c}2\\ 12\end{array}\right]\\ \left[\begin{array}{c}A\\ B\end{array}\right]={\left[\begin{array}{cc}1& 1\\ 2& -2\end{array}\right]}^{-1}\left[\begin{array}{c}2\\ 12\end{array}\right]\\ =\frac{1}{-2-2}\left[\begin{array}{cc}-2& -1\\ -2& 1\end{array}\right]\left[\begin{array}{c}2\\ 12\end{array}\right]\\ =\frac{1}{-4}\left[\begin{array}{c}-16\\ 8\end{array}\right]\\ =\left[\begin{array}{c}4\\ -2\end{array}\right]\end{array}$

The required partial fraction decomposition is $3+\frac{4}{x-2}-\frac{2}{x+2}$ .

The graph of $y=\frac{3x^2+2x}{x^2-4}$ and $y=3,y=\frac{4}{x-2}\text{ and }y=-\frac{2}{x+2}$ are shown below. It can be observed that the constant term is the horizontal asymptote of the rational function and the vertical asymptotes of the partial fractions are the same as the vertical asymptotes of the rational function.