Q. 35 - Precalculus Enhanced with Graphing Utilities

Question

In Problem, Write the partial fraction decomposition of each rational expression.
$\frac{ x-4}{x^2(x-1)}$

Step-by-Step Solution

Answer

Partial fraction decomposition of a rational expression$\frac{ x-4}{x^2(x-1)}$ is

$\frac{ x-4}{x^2(x-1)}=\frac{ 3}{x}+\frac{4}{x^2}+\frac{ -3}{(x-1)}$

Step 1. Given data

The given rational expression is 

$\frac{ x-4}{x^2(x-1)}$

Step 2. Decomposition of rational expression

Partial fraction decomposition of rational expression  

$$\begin{gathered} \frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\cdots +\frac{A_n}{x-a_n} \\ \frac{ x-4}{x^2(x-1)}=\frac{ A}{x}+\frac{B}{x^2}+\frac{ C}{(x-1)} \\ \frac{ x-4}{x^2(x-1)}=\frac{ Ax(x-1)}{x^2(x-1)}+\frac{B(x-1)}{x^2(x-1)}+\frac{C{x}^2}{x^2(x-1)} \end{gathered}$$

Step 3. Value of numerator of partial fractions

Rearrange the equation

$$\begin{gathered} \frac{ x-4}{x^2(x-1)}=\frac{ Ax(x-1)}{x^2(x-1)}+\frac{B(x-1)}{x^2(x-1)}+\frac{C{x}^2}{x^2(x-1)} \\ x-4=Ax(x-1)+B(x-1)+C{x}^2 \\ 0x^2+1x-4=(A+C)x^2+(B-A)x-B \end{gathered}$$

Equate the constants

$B=4$

Equate the coefficients of x

$$\begin{gathered} B-A=1 \\ 4-A=1 \\ A=3 \end{gathered}$$

Equate the coefficients of $x^2$

$$\begin{gathered} A+C=0 \\ 3+C=0 \\ C=-3 \end{gathered}$$

So partial fraction decomposition is  $\frac{ x-4}{x^2(x-1)}=\frac{ 3}{x}+\frac{4}{x^2}+\frac{ -3}{(x-1)}$

Step 4. Verification

Simplify the partial fractions

$$\begin{gathered} \frac{ 3}{x}+\frac{4}{x^2}+\frac{ -3}{(x-1)}=\frac{ 3x(x-1)+4(x-1)-3x^2}{x^2(x-1)} \\ =\frac{ 3x^2-3x+4x-4-3x^2}{x^2(x-1)} \\ =\frac{ x-4}{x^2(x-1)} \end{gathered}$$

It is equal to the original expression so partial fraction decomposition is correct.