Could you help me with the following problem, I don't understand how to do it.

Find an nth degree polynomial function with real coefficients satisfying the given conditions.

1. n=3; 3 and i are zeros; f(2)=20

f(x) = (x-3)(x-1)(x-a)

That function must be zero when x = 1 or 3.

Select the constant a so that f(2) = 20
(-1)(1)(2-a) = 20
a-2 = 20
a = 22

Nhbggtttitirriri

To find an nth degree polynomial function with real coefficients, we need to consider the given conditions step by step.

1. n=3: This tells us that the degree of the polynomial is 3.

2. 3 and i are zeros: Since we are given that 3 and i (which represents an imaginary number) are zeros, we know that the polynomial has the factors (x - 3) and (x - i).

3. We know that a complex number and its conjugate are both zeros of a polynomial with real coefficients. Since i is a zero, its conjugate, -i, must also be a zero. Therefore, we have (x - 3), (x - i), and (x + i) as factors.

4. f(2) = 20: Let's use this condition to determine the value of the leading coefficient in our polynomial. Substituting x = 2 into the polynomial gives:
f(2) = (2 - 3) * (2 - i) * (2 + i) * a = (2 - 3) * (2^2 - i^2) * a = -a * (4 + 1) = -a * 5
We are given that f(2) = 20, so we have:
-a * 5 = 20
Solving for a, we get: a = -4

5. Now that we have the leading coefficient, we can write the polynomial function.
The polynomial is given by:
f(x) = -4 * (x - 3) * (x - i) * (x + i)

Thus, the nth degree polynomial function with real coefficients satisfying the given conditions is f(x) = -4(x - 3)(x - i)(x + i).