Solution 2 (a).

See text and/or instructor's solution manual.

Answer.      has poles of order  3  at  ,  and a pole of order  4  at  1.

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

                    .  

Hence,      has zeros of order  3  at  ,  and a zero of order  4  at  1.

Therefore,   ,  

has poles of order  3  at  ,  and a pole of order  4  at  1.

We are  done.   

Aside.  We can let Mathematica double check our work.

We are really done.   

                    The poles of order  3  at  ,  and the pole of order  4  at  1.

                              A plot for   .  

This solution is complements of the authors.

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