How do I solve this equation?

Question

Find a polynomial of degree 3 with real coefficients and zeros of

−​3,−​1,

and​ 4, for which

​f(−​2)=18.

Answer 1

If the polynomial has zeros of -3, -1, and 4, then it has factors of (x + 3), (x + 1), and (x - 4). So the polynomial could be y = (x + 3)(x + 1)(x - 4). But it could also be y = 2(x + 3)(x + 1)(x - 4) or y = 25(x + 3)(x + 1)(x - 4). A generic way to show that is to make the polynomial be y = a(x + 3)(x + 1)(x - 4) then you can use the point (-2, 18) [from f(-2) = 18)] to calculate what the value of "a" is.

y = a(x + 3)(x + 1)(x - 4)

18 = a(-2 + 3)(-2 + 1)(-2 - 4)

18 = a(1)(-1)(-6)

18 = 6a

a = 3

So the polynomial is f(x) = 3(x + 3)(x + 1)(x - 4)

If you want, you can multiply this out. (x + 3)(x + 1) = x² + 4x + 3 and then (x² + 4x + 3)(x - 4) = x³ - 13x -12 and then (x³ - 13x - 12)(3) = 3x³ - 39x - 36 so f(x) = 3x³ - 39x - 36