Chapter 7.3, Problem 70E

Solution

To calculate : The partial fraction decomposition of $\frac{4}{\left(x+3\right)\left(x+7\right)}$ and confirm answer algebraically by combining the partial fractions.

The partial fraction decomposition of $\frac{4}{\left(x+3\right)\left(x+7\right)}$ is $X=\left[\begin{array}{c}1\\ -1\end{array}\right]$ .

Given information :

The fraction $\frac{4}{\left(x+3\right)\left(x+7\right)}$ .

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 $\frac{4}{\left(x+3\right)\left(x+7\right)}$ ,

Separate into two fractions with distinct linear denominations.

  $\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{A}{x+3}+\frac{B}{x+7}$

Multiply both sides with $\left(x+3\right)\left(x+7\right)$ .

  $4=A\left(x+7\right)+B\left(x+3\right)$

Expand the like terms.

  $\begin{array}{c}4=Ax+7A+Bx+3B\\ 4=\left(A+B\right)x+\left(7A+3B\right)\end{array}$

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

  $\begin{array}{c}A+B=0\\ 7A+3B=4\end{array}$

Write the system as matrix equation $CX=D$ where $C=\left[\begin{array}{cc}1& 1\\ 7& 3\end{array}\right]$ , $X=\left[\begin{array}{c}A\\ B\end{array}\right]$ and $D=\left[\begin{array}{c}0\\ 4\end{array}\right]$ .

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.

  

Thus, the partial fraction decomposition of $\frac{4}{\left(x+3\right)\left(x+7\right)}$ is $X=\left[\begin{array}{c}1\\ -1\end{array}\right]$ .

Now check the partial fraction decomposition by algebraically.

  $\begin{array}{c}\frac{1}{\left(x+3\right)}-\frac{1}{\left(x+7\right)}=\frac{x+7-\left(x+3\right)}{\left(x+3\right)\left(x+7\right)}\\ =\frac{x+7-x-3}{\left(x+3\right)\left(x+7\right)}\\ =\frac{4}{\left(x+3\right)\left(x+7\right)}\end{array}$

Hence, verified.