Suppose that R (x) is a polynomial of degree 7 whose coefficients are real numbers.

Also, suppose that R (x) has the following zeros.

-4 - 4i, 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 maximum number of non-real zeros that R (x) can have?

So ... it has degree of 7, so based on the fundamental theorem of algebra ... it can have at most 7 solutions

now ... you have - 4, - 4i, 2i < - - - there are two complex ones there

-4i or 0-4i and 2i or 0+2i

now ... bear in mind that, complex solutions never come all by their lonesome, they come with their sister, the conjugate

thus, that means 0-4i comes with 0+4i and 0+2i comes with 0-2i

that makes only 5 roots though

that simply means that, the - 4 one, has a multiplicity of 3

as far as the B) and C) sections, check Descartes Rule of Signs

which surely you've covered already