Question

(X-1)(x-3)(x-2)(x-6)+96

Answer 1

120=2×3×4×5=(−5)×(−4)×(−3)×(−2) When you get these roots, they turn into the estimations of an and b in the above answer.

Be careful that one of these is negative.

Partition by those two components x−a, x−b and you are left with a quadratic remainder x2+cx+d that has not any more genuine roots (on the grounds that the discriminant c2−4d is negative).