Chapter 7.3, Problem 69E

To determine

To write the Partial fraction decomposition of the rational expression $\frac{2x^3-x^2+x+5}{x^2+3x+2}$

Answer to Problem 69E

The partial fraction of expansion is $2x-7+\frac{1}{x+1}+\frac{17}{x+2}$

Explanation of Solution

Given information:

The rational expression is $\frac{2x^3-x^2+x+5}{x^2+3x+2}$

The rational expression is $\frac{2x^3-x^2+x+5}{x^2+3x+2}$

Here $N\left(x\right)/D\left(x\right)$ is an improper fraction because degree of $N\left(x\right)$

  $>$ degree of $D\left(x\right)$ so divide $N\left(x\right)$ by $D\left(x\right)$ to obtain

  $\frac{2x^3-x^2+x+5}{x^2+3x+2}=2x-7+\frac{18x+19}{x^2+3x+2}$

Now express $\frac{18x+19}{x^2+3x+2}$ in terms of partial fraction

  $\frac{18x+19}{x^2+3x+2}=\frac{A}{x+1}+\frac{B}{x+2}$

Now , for finding the value of coefficients $A$ and $B$

  $\begin{array}{l}\frac{18x+19}{x^2+3x+2}=\frac{A}{x+1}+\frac{B}{x+2}=\frac{A\left(x+2\right)+B(x+1)}{\left(x+1\right)\left(x+2\right)}=\frac{x\left(A+B\right)+\left(2A+B\right)}{\left(x+1\right)\left(x+2\right)}\\ 18x+19=x\left(A+B\right)+\left(2A+B\right)\end{array}$

Comparing coefficients of $x$

  $\begin{array}{l}A+B=18\\ B=18-A\end{array}$

Comparing constant terms in the equation

  $\begin{array}{l}2A+B=19\\ 2A+18-A=19\\ A=1\\ B=17\end{array}$

therefore, $\frac{2x^3-x^2+x+5}{x^2+3x+2}=2x-7+\frac{1}{x+1}+\frac{17}{x+2}$

Checking algebraically,

  $\begin{array}{l}2x-7+\frac{1}{x+1}+\frac{17}{x+2}=\frac{\left(2x-7\right)\left(x^2+3x+2\right)+x+2+17\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}\\ =\frac{2x^3+6x^2+4x-7x^2-21x-14+18x+19}{\left(x+1\right)\left(x+2\right)}=\frac{2x^3-x^2+x+5}{x^2+3x+2}\end{array}$

Checking Graphically,

  $\frac{2x^3-x^2+x+5}{x^2+3x+2}$

  

  $2x-7+\frac{1}{x+1}+\frac{17}{x+2}$

  

The graph of rational expression and partial fraction is same hence graphically it is proved that $\frac{2x^3-x^2+x+5}{x^2+3x+2}=2x-7+\frac{1}{x+1}+\frac{17}{x+2}$