Chapter 7, Problem 60RE

To determine

Find the partial fraction for the given rational expression.

Answer to Problem 60RE

The partial fraction form is:

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

Explanation of Solution

Given:

The rational expression

  $\frac{-x}{x^2+3x+2}$

Consider the rational expression

  $\frac{-x}{x^2+3x+2}$

Since the degree of denominator is greater than numerator, the expression is in proper form.

The denominator can be written as

  $x^2+3x+2=\left(x+2\right)\left(x+1\right)$

Take one partial fraction with constant numerator for each linear factor of denominator.

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

  $=\frac{A}{x+2}+\frac{B}{x+1}$

Simplify as follows

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

Thus

  $–x=A\left(x+1\right)+B\left(x+2\right)$ ....(1)

Substitute $x=–1$ in equation (1) to get

  $\begin{array}{l}–\left(–1\right)=A\left(–1+1\right)+B\left(–1+2\right)\hfill \\ 1=B\hfill \end{array}$

Substitute $x=–2$ in equation (1) to get

  $\begin{array}{l}\begin{array}{l}–\left(–2\right)=A\left(–2+1\right)+B\left(–2+2\right)\hfill \\ 2=–A\hfill \end{array}\\ –2=A\end{array}$

Hence, the partial fraction is given by

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