Chapter 7, Problem 52RE

Solution

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

The obtained result is $-\frac{1}{x+1}+\frac{3}{x-2}+\frac{2}{x-3}$ .

Given Information:

The rational function is defined as,

  $\frac{4x^2-3x-19}{x^3-4x^2+x+6}$

Calculation:

Consider the given rational function,

  $\frac{4x^2-3x-19}{x^3-4x^2+x+6}$

Rewrite the denominator in the factor form.

  $\frac{4x^2-3x-19}{x^3-4x^2+x+6}=\frac{-2x^2-12x+14}{\left(x+1\right)\left(x-2\right)\left(x-3\right)}$

Now, separate into two fractions with distinct linear denominators:

  $\begin{array}{c}\frac{4x^2-3x-19}{x^3-4x^2+x+6}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x-3}\\ \frac{4x^2-3x-19}{\left(x+1\right)\left(x-2\right)\left(x-3\right)}=\frac{A\left(x-2\right)\left(x-3\right)+B\left(x+1\right)\left(x-3\right)+C\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)\left(x-3\right)}\\ 4x^2-3x-19=A{x}^2-5Ax+6A+B{x}^2-2Bx-3B+C{x}^2-Cx-2C\\ 4x^2-3x-19=\left(A+B+C\right)x^2+\left(-5A-2B-C\right)x+\left(6A-3B-2C\right)\end{array}$

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

  $\begin{array}{c}A+B+C=4\\ -5A-2B-C=-3\\ 6A-3B-2C=-19\end{array}$

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

  $C=\left[\begin{array}{ccc}1& 1& 1\\ -5& -2& -1\\ 6& -3& -2\end{array}\right]$ , $X=\left[\begin{array}{c}A\\ B\\ C\end{array}\right],D=\left[\begin{array}{c}4\\ -3\\ -19\end{array}\right]$

And,

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

Find the inverse of the matrix.

  $\begin{array}{c}{C}^{-1}={\left[\begin{array}{ccc}1& 1& 1\\ -5& -2& -1\\ 6& -3& -2\end{array}\right]}^{-1}\\ =\frac{1}{\left|C\right|}\text{adj}\left(\begin{array}{ccc}1& 1& 1\\ -5& -2& -1\\ 6& -3& -2\end{array}\right)\\ =\frac{1}{12}\left(\begin{array}{ccc}1& -5& 6\\ 1& -2& -1\\ 1& -1& -2\end{array}\right)\end{array}$

And,

  $\begin{array}{c}X=C^{-1}D\\ =\frac{1}{12}\left(\begin{array}{ccc}1& -5& 6\\ 1& -2& -1\\ 1& -1& -2\end{array}\right)\cdot \left(\begin{array}{c}4\\ -3\\ -19\end{array}\right)\\ =\left(\begin{array}{c}-1\\ 3\\ 2\end{array}\right)\end{array}$

Hence, the obtained partial fraction is $-\frac{1}{x+1}+\frac{3}{x-2}+\frac{2}{x-3}$ .