To find a polynomial function of degree 3 with the given numbers as zeros and a leading coefficient of 1, we start by noting that the given zeros are [tex]\(-1\)[/tex], [tex]\(2i\)[/tex], and [tex]\(-2i\)[/tex].
Let's denote the polynomial function by [tex]\( f(x) \)[/tex]. Since the given numbers are the zeros of the polynomial, the polynomial can be written as a product of linear factors associated with each zero. Specifically, the polynomial can be written as:
[tex]\[ f(x) = (x - (-1))(x - 2i)(x + 2i) \][/tex]
[tex]\[ f(x) = (x + 1)(x - 2i)(x + 2i) \][/tex]
Next, we need to simplify this expression by first handling the complex conjugates [tex]\( (x - 2i)(x + 2i) \)[/tex]. Notice that the product of a complex number and its conjugate is always a real number:
[tex]\[ (x - 2i)(x + 2i) = x^2 - (2i)^2 \][/tex]
Since [tex]\( (2i)^2 = 4(-1) = -4 \)[/tex], we get:
[tex]\[ x^2 - (-4) = x^2 + 4 \][/tex]
Now our polynomial becomes:
[tex]\[ f(x) = (x + 1)(x^2 + 4) \][/tex]
Next, we expand this polynomial:
[tex]\[ f(x) = x(x^2 + 4) + 1(x^2 + 4) \][/tex]
[tex]\[ f(x) = x^3 + 4x + x^2 + 4 \][/tex]
Finally, rearrange the terms in descending order of exponents to obtain:
[tex]\[ f(x) = x^3 + x^2 + 4x + 4 \][/tex]
Thus, the polynomial function is:
[tex]\[ f(x) = x^3 + x^2 + 4x + 4 \][/tex]
