Chapter 7.4, Problem 91E

To determine

To find: The partial fraction decomposition using system of equations and solve using matrices.

Answer to Problem 91E

The partial fraction decomposition is $\frac{8x^2}{{(x-1)}^2(x+1)}=\frac{2}{(x+1)}+\frac{6}{(x-1)}+\frac{4}{{(x-1)}^2}$ .

Explanation of Solution

Given information: A rational expression, $\frac{8x^2}{{(x-1)}^2(x+1)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C}{{(x-1)}^2}$ .

Calculation:

Rewriting the expression, $\frac{8x^2}{{(x-1)}^2(x+1)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C}{{(x-1)}^2}$

Multiplying each sides by ${(x-1)}^2(x+1)$ , the least common denominator, leads to the basic equation,

  $\begin{array}{l}$ 8x^2=A{(x-1)}^2+B(x+1)(x-1)+C(x+1)$8x^2=A(x^2-2x+1)+B(x^2-1)+C(x+1)\\ 8x^2=(A+B)x^2+x(-2A+C)+(A-B+C)\end{array}$

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:

  $\begin{array}{l}A+B=8\\ -2A+C=0\\ A-B+C=0\end{array}$

Now the augmented matrix can be written as,

  $\left(\begin{array}{cccc}1& 1& 0& 8\\ -2& 0& 1& 0\\ 1& -1& 1& 0\end{array}\right)$

Using row transformations

  $\begin{array}{l}{R}_2\to R_2+2R_1\\ R_3\to R_3-R_1\\ \left(\begin{array}{cccc}1& 1& 0& 8\\ 0& 2& 1& 16\\ 0& -2& 1& -8\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_3\to R_3+R_2\\ \left(\begin{array}{cccc}1& 1& 0& 8\\ 0& 2& 1& 16\\ 0& 0& 2& 8\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_2\to \frac{R_2}{2}\\ \left(\begin{array}{cccc}1& 1& 0& 8\\ 0& 1& \frac{1}{2}& 8\\ 0& 0& 2& 8\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_1\to R_1-R_2\\ \left(\begin{array}{cccc}1& 0& -\frac{1}{2}& 0\\ 0& 1& \frac{1}{2}& 8\\ 0& 0& 2& 8\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_3\to \frac{1}{2}{R}_3\\ \left(\begin{array}{cccc}1& 0& -\frac{1}{2}& 0\\ 0& 1& \frac{1}{2}& 8\\ 0& 0& 1& 4\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_1\to R_1+\frac{1}{2}{R}_3\\ R_2\to R_2-\frac{1}{2}{R}_3\\ \left(\begin{array}{cccc}1& 0& 0& 2\\ 0& 1& 0& 6\\ 0& 0& 1& 4\end{array}\right)\end{array}$

Now,

  $A=2,B=6,C=4$

The partial fraction decomposition is,

  $\frac{8x^2}{{(x-1)}^2(x+1)}=\frac{2}{(x+1)}+\frac{6}{(x-1)}+\frac{4}{{(x-1)}^2}$ .