Chapter 7.4, Problem 19E

To determine

To write:The partial fraction decomposition of the rational expression $\frac{3}{x^2+x-2}$ .

Answer to Problem 19E

  $\frac{3}{\left(x-1\right)\left(x+2\right)}=\frac{1}{x-1}-\frac{1}{x+2}$

Explanation of Solution

Given:

  $\frac{3}{x^2+x-2}$

Calculation:

The given rational expression is

  $\frac{3}{x^2+x-2}$

The factor of the denominator is given by $x^2+x-2=\left(x-1\right)\left(x+2\right)$

So, the rational expression becomes $\frac{3}{\left(x-1\right)\left(x+2\right)}$

The form of partial fraction decomposition of this expression is given by

  $\frac{3}{\left(x-1\right)\left(x+2\right)}=\frac{A}{x-1}+\frac{B}{x+2}$

On simplifying

  $3=A\left(x+2\right)+B\left(x-1\right)$

Now, compare the coefficient of x ,

  $\begin{array}{l}0=A+B\\ \Rightarrow A=-B\mathrm{...}\left(i\right)\end{array}$

And comparing the constant term on both sides

  $3=2A-B$

From equation $\left(i\right)$

  $\begin{array}{c}3=2A+A\\ 3A=3\\ A=1\end{array}$

Therefore, using equation $\left(i\right)$

  $B=-1$

Therefore, the partial fraction decomposition is given by

  $\frac{3}{\left(x-1\right)\left(x+2\right)}=\frac{1}{x-1}-\frac{1}{x+2}$

In order to check the result algebraically, simplify the expression on right hand side of the equation

  $\begin{array}{c}\frac{1}{x-1}-\frac{1}{x+2}=\frac{x+2-x+1}{\left(x-1\right)\left(x+2\right)}\\ =\frac{3}{\left(x-1\right)\left(x+2\right)}\end{array}$

This resultant expression is same as the given rational expression. Hence, the partial fraction of the given expression is true.