16. Determine algebraically where the cubic polynomial function that has zeros at 2,3, and -5 and passes through the point (4,36) has a value of 120.
So I did (x-2)(x-3)(x+5)=y (4,36) SUB
(4-2)(4-3)(4+9)=36
(2)(1)(9)=36
18a/18=36/18=2
so 2(x-2)(x-3)(x+5)=120???.. im not sure what to do next.