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.