Chapter 7.4, Problem 58E

To determine

To calculate:

Write the partial fraction decomposition of the rational expression. Use a graphing utility to check your result.

  $\frac{x^3-x+3}{x^2+x-2}$

Answer to Problem 58E

The partial fraction decomposition is $\frac{x^3-x+3}{x^2+x-2}=x-1+\frac{1}{x-1}+\frac{1}{x+2}$

Explanation of Solution

Given information:

The given equation is $\frac{x^3-x+3}{x^2+x-2}$

Calculation:

Since the degree of numerator is not less than degree of the denominator we need to do polynomial long division which is as follows.

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

Now, let us deal with $\frac{2x+1}{x^2+x-2}$

Factor the denominator: $\frac{2x+1}{x^2+x-2}=\frac{2x+1}{\left(x-1\right)\left(x+2\right)}$

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction.

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

The denominators are equal, so we require the equality of the numerators:

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

Expand the right hand side:

  $2x+1=xA+xB+2A-B$

Collect up the like terms:

  $2x+1=x\left(A+B\right)+2A-B$

The coefficients near the like terms should be equal, so the following system is obtained:

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

By solving it we get, $A=1,B=1$

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

Conclusion:

The partial fraction decomposition is $\frac{x^3-x+3}{x^2+x-2}=x-1+\frac{1}{x-1}+\frac{1}{x+2}$

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