Chapter 7, Problem 59RE

To determine

Find the partial fraction for the given rational expression.

Answer to Problem 59RE

The partial fraction form is:

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

Explanation of Solution

Given:

The rational expression

  $\frac{4-x}{x^2+6x+8}$ .

Let consider the following

  $\frac{4-x}{x^2+6x+8}$

Factorize the denominator:

  $\frac{4-x}{x^2+6x+8}=\frac{4-x}{(x+4)(x+2)}$

Write in decomposition form

  $\frac{4-x}{(x+4)(x+2)}=\frac{A}{(x+4)}+\frac{B}{(x+2)}$

To find A and B, multiply the equation with $\left(x+4\right)\left(x+2\right)$ and equating co-efficient of like terms.

  $4–x=A\left(x+2\right)+B\left(x+4\right)$ Equating constant term,

  $4=2A+4B$

Dividing throughout with 2.

  $2=A+2B$

Equating co-efficient of x,

  $–2=A+B$

Solving equations,

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

Substituting back value of $B=3$ in $A+B=–1$ , get $A=–4$ .

So, $A=–4$ and $B=3$ .

The partial fraction form is:

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