Suppose that R(x) is a polynomial of degree 9 whose coefficients are real numbers. Also, suppose that R(x) has the following zeros.

-3, 6, 3i, -2+ 2i

Answer the following.

(a) Find another zero of R(x).

(b) What is the maximum number of real zeros that R(x) can have?

(c) What is the maxium number of nonreal zeros that R(x) can have?

Thank you for the opportunity to help you with your question!

if 3i is a zero then -3i is also a zero. similarly -2-2i is also a zero

-3, 6, 3i, -2+ 2i,-3i,-2-2i are the zeros

a) -3i is the other zero of R(x)

b) 4 complex as of now and a maximum of 5 real zeros R(x) can have

c) 4 complex zeros and 2 real zeros as of now. There is a possibility of two zeros being complex out of other 3 unknown zeros

So, R(x) can have a maximum of 6 complex zeros

how many zeros would there be for c? it just asks for a number

For c, a maximum number of 6 non-real zeros are possible

and a minimum of 3 real zeros

So, 3 Real zeros for c

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