Chapter 7.3, Problem 77E

Solution

To calculate : The partial fraction decomposition of $\frac{x+17}{2x^2+5x-3}$ using algebraic method.

The partial fraction decomposition of $\frac{x+17}{2x^2+5x-3}$ is $X=\left[\begin{array}{c}-2\\ 5\end{array}\right]$ .

Given information :

The fraction $\frac{x+17}{2x^2+5x-3}$ .

Formula used :

For $A$ to be coefficient matrix of a system of $n$ linear equations in $n$ variables given by $AX=B$ . If $A^{-1}$ exists, then system of equations has unique solution $X=A^{-1}B$ .

Calculation :

Consider the fraction $\frac{x+17}{2x^2+5x-3}$ ,

Factorize $2x^2+5x-3$ .

  $\begin{array}{c}2x^2+5x-3=2x^2+6x-x-3\\ =2x\left(x+3\right)-1\left(x+3\right)\\ =\left(x+3\right)\left(2x-1\right)\end{array}$

Separate into two fractions with distinct linear denominations.

  $\frac{x+17}{2x^2+5x-3}=\frac{A}{x+3}+\frac{B}{2x-1}$

Multiply both sides with $\left(x+3\right)\left(2x-1\right)$ .

  $x+17=A\left(2x-1\right)+B\left(x+3\right)$

Expand the like terms.

  $\begin{array}{c}x+17=2Ax-A+Bx+3B\\ x+17=\left(2A+B\right)x+\left(-A+3B\right)\end{array}$

Using the coefficient and constants, write the system of equations.

  $\begin{array}{c}2A+B=1\\ -A+3B=17\end{array}$

Write the system as matrix equation $CX=D$ where $C=\left[\begin{array}{cc}2& 1\\ -1& 3\end{array}\right]$ , $X=\left[\begin{array}{c}A\\ B\end{array}\right]$ and $D=\left[\begin{array}{c}1\\ 17\end{array}\right]$ .

Solve for $X$ by multiplying both sides $C^{-1}$ that is $X=C^{-1}D$ .

  $\begin{array}{c}\left[\begin{array}{cc}2& 1\\ -1& 3\end{array}\right]\left[\begin{array}{c}A\\ B\end{array}\right]=\left[\begin{array}{c}1\\ 17\end{array}\right]\\ \left[\begin{array}{c}A\\ B\end{array}\right]={\left[\begin{array}{cc}2& 1\\ -1& 3\end{array}\right]}^{-1}\left[\begin{array}{c}1\\ 17\end{array}\right]\\ \left[\begin{array}{c}A\\ B\end{array}\right]=\frac{1}{3\left(2\right)-1\left(-1\right)}\left[\begin{array}{cc}3& -1\\ 1& 2\end{array}\right]\left[\begin{array}{c}1\\ 17\end{array}\right]\\ \left[\begin{array}{c}A\\ B\end{array}\right]=\frac{1}{7}\left[\begin{array}{c}3-17\\ 1+34\end{array}\right]\\ \left[\begin{array}{c}A\\ B\end{array}\right]=\frac{1}{7}\left[\begin{array}{c}-14\\ 35\end{array}\right]\\ \left[\begin{array}{c}A\\ B\end{array}\right]=\left[\begin{array}{c}-2\\ 5\end{array}\right]\end{array}$

So, $A=-2$ and $B=5$ , the partial fraction decomposition is $\frac{-2}{x+3}+\frac{5}{2x-1}$ .

Thus, the partial fraction decomposition of $\frac{x+17}{2x^2+5x-3}$ is $\frac{-2}{x+3}+\frac{5}{2x-1}$ .