Find a polynomial function of degree 3 with the given numbers as zeros. Assume that the leading coefficient is 1 . $$ -2,4 i,-4 i $$

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To find a polynomial function of degree 3 with zeros $$ -2 $$, $$ 4i $$, and $$ -4i $$ and a leading coefficient of 1, follow these steps:

1. Since the zeros are given, the polynomial can be written in its factored form as:
   $$
   f(x) = (x + 2)(x - 4i)(x + 4i)
   $$

2. Notice that $$ (x - 4i)(x + 4i) $$ is a product of conjugates. This product can be simplified using the difference of squares:
   $$
   (x - 4i)(x + 4i) = x^2 - (4i)^2
   $$
   Since $$ (4i)^2 = -16 $$, this becomes:
   $$
   x^2 - (-16) = x^2 + 16
   $$

3. Substitute this result back into the polynomial:
   $$
   f(x) = (x + 2)(x^2 + 16)
   $$

4. Expand the product:
   $$
   f(x) = x(x^2 + 16) + 2(x^2 + 16)
   $$
   $$
   f(x) = x^3 + 16x + 2x^2 + 32
   $$

5. Arrange the terms in descending order:
   $$
   f(x) = x^3 + 2x^2 + 16x + 32
   $$

Thus, the polynomial function is:
$$
f(x) = x^3 + 2x^2 + 16x + 32
$$
The polynomial function is $$ f(x) = x^3 + 2x^2 + 16x + 32 $$.

Find a polynomial function of degree 3 with the given numbers as zeros. Assume that the leading coefficient is 1 . \( -2,4 i,-4 i \)

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