Chapter 7, Problem 51RE

Solution

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

The obtained result is $\frac{18}{x-2}-\frac{20}{x-3}$ .

Given Information:

The rational function is defined as,

  $\frac{-2x^2-12x+14}{x^3-6x^2+11x-6}$

Calculation:

Consider the given rational function,

  $\frac{-2x^2-12x+14}{x^3-6x^2+11x-6}$

Rewrite the denominator in the factor form.

  $\frac{-2x^2-12x+14}{x^3-6x^2+11x-6}=\frac{-2x^2-12x+14}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}$

Now, separate into two fractions with distinct linear denominators:

  $\begin{array}{c}\frac{-2x^2-12x+14}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\\ \frac{-2x^2-12x+14}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{A\left(x-2\right)\left(x-3\right)+B\left(x-1\right)\left(x-3\right)+C\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\\ -2x^2-12x+14=A{x}^2-5Ax+6A+B{x}^2-4Bx+3B+C{x}^2-3Cx+2C\\ -2x^2-12x+14=\left(A+B+C\right)x^2+\left(-5A-4B-3C\right)x+\left(6A+3B+2C\right)\end{array}$

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

  $\begin{array}{c}A+B+C=-2\\ -5A-4B-3C=-12\\ 6A+3B+2C=14\end{array}$

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

  $C=\left[\begin{array}{ccc}1& 1& 1\\ -5& -4& -3\\ 6& 3& 2\end{array}\right]$ , $X=\left[\begin{array}{c}A\\ B\\ C\end{array}\right],D=\left[\begin{array}{c}-2\\ -12\\ 14\end{array}\right]$

And,

  $\begin{array}{c}X=C^{-1}D\\ =\frac{1}{\left|C\right|}\text{adj}\left(C\right)D\\ =\frac{1}{2}\left(\begin{array}{ccc}1& 1& 1\\ -8& -4& -2\\ 9& 3& 1\end{array}\right)\cdot \left(\begin{array}{c}-2\\ -12\\ 14\end{array}\right)\\ =\left(\begin{array}{c}0\\ 18\\ -20\end{array}\right)\end{array}$

Hence, the obtained partial fraction is $\frac{18}{x-2}-\frac{20}{x-3}$ .