Chapter 7.4, Problem 93E

To determine

To find: The values of A, B, C, D by solving the system of equations using matrices.

Answer to Problem 93E

The values are,

  $A=-1,B=3,C=2,D=-3$

And the partial fraction decomposition is

  $\frac{x^3-2x+6}{x^2(x^2+2)}=\frac{-1}{x}+\frac{3}{x^2}+\frac{2x-3}{(x^2+2)}$ .

Explanation of Solution

Given information: A rational expression, $\frac{x^3-2x+6}{x^2(x^2+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{(x^2+2)}$

Calculation:

Rewriting the expression, $\frac{x^3-2x+6}{x^2(x^2+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{(x^2+2)}$

Multiplying each side by $x^2(x^2+2)$ , the least common denominator, leads to the basic equation,

  $\begin{array}{l}{x}^3-2x+6=Ax(x^2+2)+B(x^2+2)+(Cx+D)x^2\\ x^3-2x+6=x^3(A+C)+x^2(B+D)+x(2A)+(2B)\end{array}$

Two polynomials are equal if and only if the coefficients of like terms are equal. Equating the coefficients we can write the system of linear equations as:

  $\begin{array}{l}A+C=1\\ B+D=0\\ 2A=-2\\ 2B=6\end{array}$

Now the augmented matrix can be written as,

  $\left(\begin{array}{ccccc}1& 0& 1& 0& 1\\ 0& 1& 0& 1& 0\\ 2& 0& 0& 0& -2\\ 0& 2& 0& 0& 6\end{array}\right)$

Using row transformations

  $\begin{array}{l}{R}_3\to R_3-2R_1\\ R_4\to R_4-2R_2\\ \left(\begin{array}{ccccc}1& 0& 1& 0& 1\\ 0& 1& 0& 1& 0\\ 0& 0& -2& 0& -4\\ 0& 0& 0& -2& 6\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_3\to -\frac{1}{2}{R}_3\\ R_4\to -\frac{1}{2}{R}_4\\ \left(\begin{array}{ccccc}1& 0& 1& 0& 1\\ 0& 1& 0& 1& 0\\ 0& 0& 1& 0& 2\\ 0& 0& 0& 1& -3\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_1\to R_1-R_3\\ R_2\to R_2-R_4\\ \left(\begin{array}{ccccc}1& 0& 0& 0& -1\\ 0& 1& 0& 0& 3\\ 0& 0& 1& 0& 2\\ 0& 0& 0& 1& -3\end{array}\right)\end{array}$

Now,

  $A=-1,B=3,C=2,D=-3$

The partial fraction decomposition is,

  $\frac{x^3-2x+6}{x^2(x^2+2)}=\frac{-1}{x}+\frac{3}{x^2}+\frac{2x-3}{(x^2+2)}$