Let's divide:
by using synthetic division:
We must express as
c | 3 0 1 0 5
| 3c 3c2 c+3c3 c2+3c4
----------------------------------------
3 3c 1+3c2 c+3c3 5+c2+3c4
The remainder is 5+c2+3c4. This is always
a positive number because even powers of a
real number are never negative. Thus this
remainder can never be 0.
Edwin