Question

I have attempted the question so many times and cannot come up with the right answer, I really need help.
form a polynomial f(x) with real coefficients having the given degree and zeros.degree 4 zeros 5+3i; -4 multiplicity 2

Answer 1

You need to know a few things to come up with the right answer.

Answer 2

1. For any solution k there is, there is a binomial of the form (x - k)

Answer 3

2. Multiplicity 2 means there are two of those solutions.
That means -4 multiplicity 2 means there are two solutions of -4 which means
(x - (-4) ) and (x - (-4) ) again. Simplifying, you get (x + 4)(x + 4)

Answer 4

3. A polynomial equation with real coefficients only has complex solutions in complex conjugate pairs.
That means if 5 + 3i is a solution, then its complex conjugate, 5 - 3i, must also be a solution.

Answer 5

Do you understand this so far?

Answer 6

yes

Answer 7

The polynomial is of the form
\((x - k_1)(x - k_2)...(x - k_n)\)

Since there are 4 zeros, there are 4 binomials that need to be multiplied together.
The polynomial is
\(f(x) = [x - (-4)][x - (-4)][x - (5 + 3i)][x - (5 - 3i)] \)

Now you need to simplify it.

Answer 8

Thank you I will work my way through

Answer 9

You're welcome.
You need to simplify each binomial.
Then multiply them together.