Question

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

Answer 1

start with
\[f(x)=a(x-1)(x-2)(x-3)\]

Answer 2

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

Answer 3

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

Answer 4

i get \(-6a=-18\) solve for \(a\) and you are done

Answer 5

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?

Answer 6

i dont think it's c...

Answer 7

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

Answer 8

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