What is the third-degree polynomial function such that f(0) = -18 and whose zeros are 1, 2, and 3

Question

anonymous · Answer

start with 
$$f(x)=a(x-1)(x-2)(x-3)$$

anonymous · Answer

that will give you the zeros you want. now you need to solve for $a$ and you do that by putting $$f(0)=18=a(0-1)(0-2)(0-3)$$

anonymous · Answer

ok my fault it is $f(0)=-18$ so put 
$$f(0)=-18=a(0-1)(0-2)(0-3)$$

anonymous · Answer

i get $-6a=-18$ solve for $a$ and you are done

anonymous · Answer

a      f(x) = 2x3 - 12x2 + 18x - 18
b	f(x) = 3x3 + 18x2 - 33x - 18
c	f(x) = 2x3 + 12x2 - 18x - 18
d	f(x) = 3x3 - 18x2 + 33x - 18

are my options.

i got c but i'm not sure?

lgbasallote · Answer

i dont think it's c...

lgbasallote · Answer

what did you get for (x-1)(x-2)(x-3)?

anonymous · Answer

i'm really not sure what i did. would it be letter ...... a?