Find an nth-degree polynomial function with real coefficients, a leading coefficient of 1, and satisfying the given conditions. n=3; 4 and 4i are zeros

Question

Find an nth-degree polynomial function with real coefficients, a leading coefficient of 1, and satisfying the given conditions.

n=3; 4 and 4i are zeros

Answer 1

f(x) = (x - 4)(x - 4i)(x - (-4i)) = (x - 4)(x² + 16) = x³ - 4x² + 16x - 64

Answer 2

If 4 and 4i are zeros of the polynomial function, then their conjugates -4i and 4 are also zeros, since complex zeros of a polynomial with real coefficients always come in conjugate pairs. Thus, the polynomial has four zeros: 4, -4i, 4i, and some unknown real number a.

Since the polynomial has degree 3 and a leading coefficient of 1, it must have the form:

f(x) = (x - 4)(x - 4i)(x + 4i)(x - a)

Multiplying out the factors, we get:

f(x) = (x - 4)(x^2 + 16)(x - a)

Expanding this expression, we get:

f(x) = x^4 - (4 + a)*x^3 + (16a + 64)*x^2 - 64ax + 64a

Since the polynomial has real coefficients, the imaginary parts of its complex zeros must cancel out. That is, the sum of the roots -4i and 4i must be 0, which means that the coefficient of x in f(x) must be 0:

-64a = 0

Solving for a, we get:

a = 0

Now, substituting a = 0 into the expression for f(x), we get:

f(x) = x^4 - 4x^3 + 64x^2

Therefore, the nth-degree polynomial function with real coefficients, a leading coefficient of 1, and zeros 4, 4i, and -4i is:

f(x) = x^4 - 4x^3 + 64x^2.