Chapter 7, Problem 50RE

Solution

To calculate: The partial fraction decomposition of the rational function.

The obtained result is $-\frac{5}{\left(x+2\right)}+\frac{3}{\left(x+4\right)}$ .

Given Information:

The rational function is defined as,

  $\frac{-2x-14}{x^2+6x+8}$

Calculation:

Consider the given rational function,

  $\frac{-2x-14}{x^2+6x+8}$

Rewrite the denominator in the factor form.

  $\frac{-2x-14}{x^2+6x+8}=\frac{-2x-14}{\left(x+2\right)\left(x+4\right)}$

Now, separate into two fractions with distinct linear denominators:

  $\begin{array}{c}\frac{-2x-14}{\left(x+2\right)\left(x+4\right)}=\frac{A}{\left(x+2\right)}+\frac{B}{\left(x+4\right)}\\ \frac{-2x-14}{\left(x+2\right)\left(x+4\right)}=\frac{A\left(x+4\right)+B\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}\\ -2x-14=Ax+Bx+4A+2B\\ -2x-14=x\left(A+B\right)+\left(4A+2B\right)\end{array}$

Using the coefficients and constants, we can write the system of equations:

  $\begin{array}{c}A+B=-2\\ 4A+2B=-14\end{array}$

Write the above equation as matrix system as $CX=D$ .

  $C=\left[\begin{array}{cc}1& 1\\ 4& 2\end{array}\right]$ , $X=\left[\begin{array}{c}A\\ B\end{array}\right],C=\left[\begin{array}{c}-2\\ -14\end{array}\right]$

And,

  $\begin{array}{c}{C}^{-1}CX=C^{-1}D\\ X=C^{-1}D\end{array}$

Find the inverse of the matrix.

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

And,

  $\begin{array}{c}{C}^{-1}D=-\frac{1}{2}\left[\begin{array}{cc}2& -1\\ -4& 1\end{array}\right]\cdot \left[\begin{array}{c}-2\\ -14\end{array}\right]\\ =\left[\begin{array}{c}-5\\ 3\end{array}\right]\end{array}$

Hence, the obtained partial fraction is $-\frac{5}{\left(x+2\right)}+\frac{3}{\left(x+4\right)}$ .