Solution 2 (c).

See text and/or instructor's solution manual.

Answer.      has simple poles at  ,  ,  and  .  

Solution.   Use a convenient method to determine where the denominator is zero.   As in Exercise 1 (g), we can factor the denominator and get  

                    ,  
                    
Hence,      has six simple zeros at   ,  ,   and   .  
                    
Therefore,   ,  

has six 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