Chapter 11, Problem 59RE

To determine

Write the partial fraction decomposition of each rational expression.

Answer to Problem 59RE

  $\boxed{\frac{-\frac{3}{2}}{x}+\frac{\frac{3}{2}}{x-4}}$

Explanation of Solution

Given information:

Write the partial fraction decomposition of each rational expression.

  $\frac{6}{x\left(x-4\right)}$

Calculation:

The denominator is already in factored from and it contains only non repeated linear factors.

We can decompose the given expression according to the equation,

  $\frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\mathrm{........}+\frac{A_n}{x-a_n}$

So, $\frac{6}{x\left(x-4\right)}=\frac{A}{x}+\frac{B}{x-4}$

To find $A$ and $B$ clear the fractions by multiplying each side of the equation by $x\left(x-4\right)$ .

  $6=A\left(x-4\right)+Bx$

This equation is an identify in $x$ . Equate the coefficients of $x$ in this equation.

  $0=A+B$

Equate the constants.

  $6=-4A$

We get the system of equations as $\{\begin{array}{l}0=A+B\\ 6=-4A\end{array}$ .

Substitute $-\frac{3}{2}$ for $A$ in the equation $0=A+B$ .

  $B=\frac{3}{2}$

Substitute $-\frac{3}{2}$ for $A$ and $\frac{3}{2}$ for $B$ in the equation $\frac{6}{x\left(x-4\right)}=\frac{A}{x}+\frac{B}{x-4}$ .

  $\frac{6}{x\left(x-4\right)}=\frac{-\frac{3}{2}}{x}+\frac{\frac{3}{2}}{x-4}$

Hence, the partial fraction decomposition of $\frac{6}{x\left(x-4\right)}is\frac{-\frac{3}{2}}{x}+\frac{\frac{3}{2}}{x-4}$

For checking the decomposition, add the rational expression.

  $\frac{-\frac{3}{2}}{x}+\frac{\frac{3}{2}}{x-4}=-\frac{3}{2}\left[\frac{1}{x}-\frac{1}{x-4}\right]$

  $=-\frac{3}{2}\left[\frac{x-4-x}{x\left(x-4\right)}\right]$

  $=-\frac{3}{2}\left[\frac{-4}{x\left(x-4\right)}\right]$

  $=\frac{6}{x\left(x-4\right)}$

Hence, the partial fraction decomposition is $\boxed{\frac{-\frac{3}{2}}{x}+\frac{\frac{3}{2}}{x-4}}$ .