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 algebrait 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

300b(x) is graphed by the functio