Answer:
If a cubic function has a zero at 4, we know that it has a factor of (x - 4). If it has a zero at 2i, we know that it also has a factor of (x - 2i) and another factor of (x + 2i) (since complex zeros always come in conjugate pairs).
So, the cubic function in standard form would be:
f(x) = a(x - 4)(x - 2i)(x + 2i)
Multiplying out the factors, we get:
f(x) = a(x - 4)(x^2 + 4)
Expanding further, we get:
f(x) = ax^3 - 4ax^2 + 4ax - 16a
Therefore, a cubic function in standard form that has zeros at 4 and 2i is:
f(x) = ax^3 - 4ax^2 + 4ax - 16a