Question

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

Answer 1

`f(x) = 3x^3-18x^2+33x-18`

Explanation:

To set the zeros of the function, we can first write the function as

`f(x) = (x-1)(x-2)(x-3)`

This gives us `f(1) = f(2) = f(3) = 0`, however `f(0) = -1*-2*-3 = -6`

To fix this, we can multiply by a constant.
`-18 = 3*-6` so we multiply by `3` to obtain

`f(x) = 3(x-1)(x-2)(x-3)`

which has the desired properties. Multiplying that out gives us the cubic function

`f(x) = 3x^3-18x^2+33x-18`