Chapter 10.7, Problem 18E

To determine

The partial decomposition of the given rational function is to be estimated:

  $\frac{x-12}{x^2-4x}$

Answer to Problem 18E

The partial decomposition of $\frac{x-12}{x^2-4x}$ is to be estimated as: $\frac{3}{x}+\frac{-2}{x-4}$

Explanation of Solution

Given:

  $\frac{x-12}{x^2-4x}$

Concept Used:

The process of starting the given rational number with simplified answer steps is the partial-fraction decomposition.

Calculation:

The form of the partial decomposition is expressed by the following expression:

  $\begin{array}{l}\frac{x-12}{x^2-4x}=\frac{x-12}{x\left(x-4\right)}\\ \frac{x-12}{x^2-4x}=\frac{A}{x}+\frac{B}{x-4}\end{array}$

Now, write the right hand side of the equation into a single fraction:

  $\frac{x-12}{x^2-4x}=\frac{xB+\left(x-4\right)A}{x\left(x-4\right)}$

As the denominators of both eqns. are equal so use the equality of numerators:

  $\begin{array}{l}x-12=xB+\left(x-4\right)A\\ x-12=xA+xB-4A\\ x-12=x\left(A+B\right)-4A\end{array}$

The coefficients near the like terms should be equal, so the following system is obtained:

  $A+B=1\mathrm{...}\left(1\right)$

And

  $\begin{array}{l}-4A=-12\\ A=-\frac{12}{4}=3\mathrm{...}\left(2\right)\end{array}$

Now to get value of $B$ , put the value of $A=3$ in eq. (1):

  $\begin{array}{l}A+B=1\\ 3+B=1\\ B=-2\end{array}$

Therefore, the partial decomposition of the given rational function is to be estimated as:

  $\begin{array}{l}\frac{x-12}{x^2-4x}=\frac{A}{x}+\frac{B}{x-4}\\ \frac{x-12}{x^2-4x}=\frac{3}{x}+\frac{-2}{x-4}\end{array}$

Conclusion:

Hence, $\frac{x-12}{x^2-4x}=\frac{3}{x}+\frac{-2}{x-4}$