Chapter 7.3, Problem 74E

Solution

To calculate : The partial fraction decomposition of $\frac{-6}{x^2-3x}$ and support answer graphically.

The partial fraction decomposition of $\frac{-6}{x^2-3x}$ is $\frac{2}{x}+\frac{-2}{x-3}$ .

Given information :

The fraction $\frac{-6}{x^2-3x}$ .

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 $\frac{-6}{x^2-3x}$ ,

Separate into two fractions with distinct linear denominations.

  $\frac{-6}{x^2-3x}=\frac{A}{x}+\frac{B}{x-3}$

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

  $-6=A\left(x-3\right)+B\left(x\right)$

Expand the like terms.

  $\begin{array}{c}-6=Ax-3A+Bx\\ -6=\left(A+B\right)x-3A\end{array}$

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

  $\begin{array}{c}A+B=0\\ -3A=-6\end{array}$

Write the system as matrix equation $CX=D$ where $C=\left[\begin{array}{cc}1& 1\\ -3& 0\end{array}\right]$ , $X=\left[\begin{array}{c}A\\ B\end{array}\right]$ and $D=\left[\begin{array}{c}0\\ -6\end{array}\right]$ .

Solve for $X$ by multiplying both sides $C^{-1}$ that is $X=C^{-1}D$ . Use graphing calculator to find the solution of the system as shown below.

  

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

Graph the functions $y_1=\frac{-6}{x^2-3x}$ and $y_2=\frac{2}{x}+\frac{-2}{x-3}$ .

  

Thus, the partial fraction decomposition of $\frac{-6}{x^2-3x}$ is $\frac{2}{x}+\frac{-2}{x-3}$ .