Find the polynomial function f with real coefficients that has the given degree, zeros, and solution

Question

Degree: 4; Zeros: -3, 1, i; Solution point: f(0)= -12

Answer 1

A couple of key items:

1) A degree 4 polynomial will have 4 total zeros (real + imaginary/complex)

2) Complex zeros always come in conjugate pairs. So if "i" is a zero then "-i" will also be a zero.

3) If "-3" is a zero then "x - -3" (or "x + 3") is a factor.

So there are 4 total zeros and they are -3, 1, i, -i and the factors are (x + 3), (x - 1), (x + i), (x - i)

You can have any multiplier in front of the polynomial. We can use "a" as that multiplier. So the polynomial is:

f(x) = a(x + 3)(x - 1)(x + i)(x - i)

To determine the value of "a", you can plug in the point (0, -12) and solve. Before you do that, at least multiply (x + i)(x - i) together to eliminate the imaginary numbers.