Chapter 7.3, Problem 75E

Solution

To calculate : The partial fraction decomposition of $\frac{-x+10}{x^2+x-12}$ and support answer graphically.

The partial fraction decomposition of $\frac{-x+10}{x^2+x-12}$ is $\frac{1}{x-3}+\frac{-2}{x+4}$ .

Given information :

The fraction $\frac{-x+10}{x^2+x-12}$ .

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 $\frac{-x+10}{x^2+x-12}$ ,

Factorize $x^2+x-12$ .

  $\begin{array}{c}{x}^2+x-12=x^2+4x-3x-12\\ =x\left(x+4\right)-3\left(x+4\right)\\ =\left(x-3\right)\left(x+4\right)\end{array}$

Separate into two fractions with distinct linear denominations.

  $\frac{-x+10}{x^2+x-12}=\frac{A}{x-3}+\frac{B}{x+4}$

Multiply both sides with $\left(x-3\right)\left(x+4\right)$ .

  $-x+10=A\left(x+4\right)+B\left(x-3\right)$

Expand the like terms.

  $\begin{array}{c}-x+10=Ax+4A+Bx-3B\\ -x+10=\left(A+B\right)x+\left(4A-3B\right)\end{array}$

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

  $\begin{array}{c}A+B=-1\\ 4A-3B=10\end{array}$

Write the system as matrix equation $CX=D$ where $C=\left[\begin{array}{cc}1& 1\\ 4& -3\end{array}\right]$ , $X=\left[\begin{array}{c}A\\ B\end{array}\right]$ and $D=\left[\begin{array}{c}1\\ -10\end{array}\right]$ .

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, $A=1$ and $B=-2$ , the partial fraction decomposition is $\frac{1}{x-3}+\frac{-2}{x+4}$ .

Graph the functions $y_1=\frac{-x+10}{x^2+x-12}$ and $y_2=\frac{1}{x-3}+\frac{-2}{x+4}$ .

  

Thus, the partial fraction decomposition of $\frac{-x+10}{x^2+x-12}$ is $\frac{1}{x-3}+\frac{-2}{x+4}$ .