Solution 2 (e).

See text and/or instructor's solution manual.

Answer.      has four simple poles at    and  .

Solution.   Use a convenient method to determine where the denominator is zero.   For example, we can factor the denominator and get  

                      

Hence,      has four simple zeros at    and  .   

Therefore,      has four simple poles at    and  .  

We are  done.   

Aside.  We can let Mathematica double check our work.

We are really done.   

                    The simple poles at    and  .

                              A plot for   .  

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell