Q. 48 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 47–54, use the division algorithm to rewrite each improper fraction as the sum of a quotient and proper fraction. Find the partial fraction decomposition of the proper fraction. Finally, express the improper fraction as the sum of a quotient and the partial  fraction decomposition

   $\frac{x^3 - 3x^2+ 1}{x^2+ 5x + 6}$

Step-by-Step Solution

Answer

$\frac{x^3 - 3x^2+ 1}{x^2+ 5x + 6}=(x-8)+\frac{53}{x+3}-\frac{19}{x+2}$

Step 1: Given

Improper fraction $\frac{x^3 - 3x^2+ 1}{x^2+ 5x + 6}$

Step 2: Express in partial fraction decomposition.

First, we have to convert the improper fraction as the sum of a quotient and a proper fraction. 

$\frac{x^3 - 3x^2+ 1}{x^2+ 5x + 6}=(x-8)+\frac{34x+49}{(x+3)(x+2)}$

Now, find the partial fraction of the second term.

$$\begin{gathered} \frac{34x+49}{(x+3)(x+2)}=\frac{A}{x+3}+\frac{B}{x+2} \\ 34x+49=A(x+2)+B(x+3) \end{gathered}$$

After comparing the coefficient of variable and constant.

We get that,

$$\begin{gathered} A+B=34 \\ 2A+3B=49 \end{gathered}$$

By using the substitution method,

$$\begin{gathered} 2A+3(34-A)=49 \\ 2A+102-3A=49 \\ -A=-53 \\ A=53 \end{gathered}$$

So that, $B=34-53=-19$

Therefore,

$\frac{34x+49}{(x+3)(x+2)}=\frac{53}{x+3}-\frac{19}{x+2}$

So that, given improper fraction can be written as,

$\frac{x^3 - 3x^2+ 1}{x^2+ 5x + 6}=(x-8)+\frac{53}{x+3}-\frac{19}{x+2}$