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