Find the polynomial function

[Guest]

Find the polynomial function f with real coefficients that has the given degree, zeros, and solution point.
Degree: 3
Zeros: -3, 1+[tex]\sqrt{3}i[/tex]
Solution Point: f(−2) = 12

[HallsofIvy]

Any cubic function can be written f(x)= a(x- b)(x- c)(x- d) where b, c, and d are zeros of the function. Since the coefficients are required to be real and [tex]1+ i\sqrt{3}[/tex] is a root, its complex conjugate, [tex]1- i\sqrt{3}[/tex] is also. A cubic function with those zeros is of the form [tex]f(x)= a(x+ 3)(x-1-i\sqrt{3})(x-1+\sqrt{3})[/tex].

The fact that f(-2)= 12 means that [tex]f(-2)= a(-2+ 3)(-2-1-i\sqrt{3})(-2-1+\sqrt{3})= a(-3- i\sqrt{3})(-3+i\sqrt{3})= 12[/tex]. Solve that equation for a.

[GeradHum]

Use the zeros to write f(x) = a(x + 3)(x - (1 + 3i))(x - (1 - 3i)). Then expand and use f(−2) = 12 to solve for a.