Math 2 Sat II Question

ax^5+bx^4+cx^3+dx^2+e=0

Let a, b, c, d and e represent nonzero real numbers in the equation above. If the equation has 2i as a root, which of the following statements must be true?
(A) The only other nonreal root of the equation is -2i.
(B) The equation has an odd number of nonreal roots.
© The equation has exactly one real root.
(D) The equation has an odd number of real roots.
(E) All real roots of the equation are positive.

The correct answer is D. I know that there are 5 roots because of the first coefficient’s exponent but how do I know whether those roots are real, complex, etc.?

@chelsalina There is a theorem in algebra that says that if x is a complex root of a polynomial with real coefficients, then so is xbar where xbar is the complex conjugate of x (you can see a proof on [Wikipedia](https://en.wikipedia.org/wiki/Complex_conjugate_root_theorem#Simple_proof)). If x = 2i is a root, then so is x = -2i.

(A) is not true because, while -2i is necessarily a root, it doesn’t have to be the only other non-real root. For example the polynomial could be written in the form (x-2i)(x+2i)(x-i)(x+i)(x-r).

(B) is not true - given a polynomial with real coefficients, the number of nonreal roots is necessarily even.

(C) is not true, the polynomial could have three real roots (e.g. (x-2i)(x+2i)(x-1)(x-2)(x-3) has real roots 1, 2, 3)

(D) is true but only if we count multiple roots as being distinct. Because the polynomial has five roots including multiple roots and an even number of them are nonreal, an odd number of them are real.

(E) is not true, x+1 could be a factor which makes -1 a root.

Could this realistically come up a s a question this may or june for sat 2 maht?http://talk.collegeconfidential.com/discussion/1882760/math-2-sat-ii-question?#

@happywharton You likely won’t see this exact same question, but I’d imagine similar questions about complex roots of polynomials can show up.

~found the answer to the question~

@chelsalina It’s D (you said it in your original post, right?).

I posted another question but now I understand it, anyways thanks for answering the original question!

Ah okay.

Yeah no problem!