To find a polynomial function of degree 3 with the given zeros [tex]\(-4\)[/tex], [tex]\(8i\)[/tex], and [tex]\(-8i\)[/tex], we can start by using the fact that if [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the roots (or zeros) of the polynomial, then the polynomial can be constructed as:
[tex]\( f(x) = (x-a)(x-b)(x-c) \)[/tex]
Here, the zeros are [tex]\( -4 \)[/tex], [tex]\( 8i \)[/tex], and [tex]\( -8i \)[/tex]. Therefore, the polynomial function is:
[tex]\[ f(x) = (x + 4)(x - 8i)(x + 8i) \][/tex]
Next, we simplify the expression step-by-step.
First, notice that [tex]\( (x - 8i)(x + 8i) \)[/tex] resembles the difference of squares formula, [tex]\( (a - b)(a + b) = a^2 - b^2 \)[/tex]:
[tex]\[ (x - 8i)(x + 8i) = x^2 - (8i)^2 \][/tex]
Since [tex]\( (8i)^2 = 64i^2 \)[/tex] and knowing that [tex]\( i^2 = -1 \)[/tex]:
[tex]\[ (8i)^2 = 64 \cdot (-1) = -64 \][/tex]
Therefore:
[tex]\[ x^2 - (8i)^2 = x^2 - (-64) = x^2 + 64 \][/tex]
This simplifies our polynomial to:
[tex]\[ f(x) = (x + 4)(x^2 + 64) \][/tex]
Now, distribute [tex]\( x + 4 \)[/tex] across [tex]\( x^2 + 64 \)[/tex]:
[tex]\[ f(x) = x(x^2 + 64) + 4(x^2 + 64) \][/tex]
[tex]\[ f(x) = x^3 + 64x + 4x^2 + 4 \cdot 64 \][/tex]
[tex]\[ f(x) = x^3 + 4x^2 + 64x + 256 \][/tex]
Thus, the polynomial function, simplified, is:
[tex]\[ f(x) = x^3 + 4x^2 + 64x + 256 \][/tex]
So, the polynomial function of degree 3 with the given zeros [tex]\(-4\)[/tex], [tex]\(8i\)[/tex], and [tex]\(-8i\)[/tex], and a leading coefficient of 1, is:
[tex]\[
\boxed{x^3 + 4x^2 + 64x + 256}
\][/tex]
