Question

What is the third–degree polynomial function such that f(0) = –24 and whose zeros are 1, 2, and 3?

f(x) = 4x^3 – 24x^2 + 44x – 24

f(x) = 2x^3 – 24x^2 + 22x – 24

f(x) = 4x^3 + 24x^2 – 44x – 24

f(x) = 2x^3 + 24x^2 – 22x – 24

Answer 1

If you know what the zeroes should be, this tells you what the factored form should look like. Knowing for example that -1 is a root, you know that (x + 1) should be a factor. From the givens, the three factors should be

x + 1, x - 2, and x + 3.

I hope this is clear. The cubic can have any leading coefficient---we choose the one that forces f(0) = -24. So in general

f(x) = a(x + 1)(x - 2)(x + 3).

This means that

f(0) = a(1)(-2)(3) = -24 ==> -6a = -24.

Choosing a = 4 gives the correct y-intercept. So the polynomial is

f(x) = 4(x + 1)(x - 2)(x + 3) = 4x^3 + 8x² - 20x - 24.