I have two questions that I don't understand and need help with.

1. information is given about a polynomial f(x) whose coefficients are real numbers. Find the remaining zerosof f.

degree 4, zeros i;9+i

2. form a polynomial f(x) with real coefficients having the given degree and zeros.

degree 4, zeros 3, multiplicity 2; 6i

f(x)=a(?)

1.

Need four zeros but every complex root comes with its complex conjugate
i so also -i
9+i so also 9-i

(x-3)(x-3)(x- 6i)(x+6i)
multiply that out

1. To find the remaining zeros of the polynomial f(x), we need to use the fact that complex zeros of polynomials with real coefficients always occur in conjugate pairs.

Given that the polynomial f(x) has a degree of 4 and the zeros i and 9+i, we know that i is a zero, which means its conjugate -i is also a zero. So, the remaining zeros will be -i and its conjugate 9-i.

Therefore, the complete set of zeros for f(x) will be: i, -i, 9+i, and 9-i.

2. To form a polynomial f(x) with real coefficients using the given zeros, we need to consider the fact that complex zeros always come in conjugate pairs.

Given that the polynomial f(x) has a degree of 4 and the zeros 3 (with multiplicity 2) and 6i, we know that 3 will have a multiplicity of 2, and 6i implies that its conjugate -6i will also be a zero.

To form the polynomial, we can start by knowing that (x - a) represents a linear factor of the polynomial, where 'a' is a zero. Since 3 has a multiplicity of 2, it means that the factor (x - 3) will appear twice. Similarly, (x - 6i) and (x + 6i) will form a quadratic factor, as they are conjugate pairs.

So the polynomial f(x) can be written as:
f(x) = (x - 3)(x - 3)(x - 6i)(x + 6i)

If you want to simplify the polynomial further, you can expand this using algebraic methods, which would involve multiplying the factors together to get the complete polynomial expression.