There are three possibities with the coefficient of x4 = 1 1. x=-4 has multiplicity 1 and x=-1 has multiplicity 3 f(x) = (x+4)(x+1)³ f(x) = (x+4)(x+1)(x+1)(x+1) Multiply that out and get: f(x) = x4 + 7x3 + 15x2 + 13x + 4 2. x=-4 has multiplicity 2 and x=-1 has multiplicity 2 f(x) = (x+4)²(x+1)² f(x) = (x+4)(x+4)(x+1)(x+1) Multiply that out and get: f(x) = x4 + 10x3 + 33x2 + 40x + 16 3. x=-4 has multiplicity 3 and x=-1 has multiplicity 1 f(x) = (x+4)³(x+1) f(x) = (x+4)(x+4)(x+4)(x+1) Multiply that out and get: f(x) = x4 + 13x3 + 60x2 + 112x + 64 Edwin