Chapter 7, Problem 48RE

Solution

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

The obtained result is $\frac{6}{\left(x+2\right)}-\frac{5}{\left(x-1\right)}$ .

Given Information:

The rational function is defined as,

  $\frac{x-16}{x^2+x-2}$

Calculation:

Consider the given rational function,

  $\frac{x-16}{x^2+x-2}$

Rewrite the denominator in the factor form.

  $\frac{x-16}{x^2+x-2}=\frac{x-16}{\left(x+2\right)\left(x-1\right)}$

Now, separate into two fractions with distinct linear denominators:

  $\begin{array}{c}\frac{x-16}{\left(x+2\right)\left(x-1\right)}=\frac{A}{\left(x+2\right)}+\frac{B}{\left(x-1\right)}\\ \frac{x-16}{\left(x+2\right)\left(x-1\right)}=\frac{A\left(x-1\right)+B\left(x+2\right)}{\left(x+2\right)\left(x-1\right)}\\ x-16=Ax+Bx-A+2B\\ x-16=x\left(A+B\right)+\left(-A+2B\right)\end{array}$

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

  $\begin{array}{c}A+B=1\\ -A+2B=-16\end{array}$

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

  $C=\left[\begin{array}{cc}1& 1\\ -1& 2\end{array}\right]$ , $X=\left[\begin{array}{c}A\\ B\end{array}\right],C=\left[\begin{array}{c}1\\ -16\end{array}\right]$

And,

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

Find the inverse of the matrix.

  $\begin{array}{c}{\left[\begin{array}{cc}1& 1\\ -1& 2\end{array}\right]}^{-1}=\frac{1}{2+1}\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]\\ =\frac{1}{3}\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]\end{array}$

And,

  $\begin{array}{c}{C}^{-1}D=\frac{1}{3}\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]\cdot \left[\begin{array}{c}1\\ -16\end{array}\right]\\ =\left[\begin{array}{c}6\\ -5\end{array}\right]\end{array}$

Hence, the obtained partial fraction is $\frac{6}{\left(x+2\right)}-\frac{5}{\left(x-1\right)}$ .