It is known that the two zeros of the function f (x) = AX2 + (B-8) x-a-ab are - 3 and 2 respectively. (I) find f (x); (II) find the range of F (x) when the domain of F (x) is [0,1]

It is known that the two zeros of the function f (x) = AX2 + (B-8) x-a-ab are - 3 and 2 respectively. (I) find f (x); (II) find the range of F (x) when the domain of F (x) is [0,1]

(1) From the problem meaning - B-8A = - 3 + 2-A (1 + b) a = - 3 × 2, the solution is a = - 3B = 5, | f (x) = - 3x2-3x + 18. (II) f (x) = - 3 (x + 12) 2 + 34 + 18, in [0,1] monotone decreasing, | f (1) = 12 ≤ f (x) ≤ f (0) = 18, the value range of the function is [12,18]