polynomial with complex roots....

Right. This is really getting on my nerves.

The question is:

Let $\displaystyle f=a_{n}X^n+a_{n-1}X^{n-1}+.........+a_0$ be a polynomial where the coefficientsn$\displaystyle a_0,.......,a_n$ are real and $\displaystyle a_n \neq 0.$

i). Show that if $\displaystyle \alpha$ is a complex root of f then it's conjugate $\displaystyle \overline {\alpha} $ is also a root of f.

Would this just be that the product of a complex number and it's conjugate produce a real number. Since this polynomial has real coefficients then there can't be a complex number on it's own (or some of the coefficients would be complex)?

ii) Suppose that the degree n is odd. Show that f must have at least one real root.

Would the solution to this be that a polynomial with a degree n has n roots. If the polynomial is odd then there would be $\displaystyle \frac {n-1}{2}$ pairs of complex numbers and their conjugates and one root left over. Since this root is on it's own, it must be real to avoid complex coefficients.

I do not know how to prove this though (Crying).

Please help!

P.S: Sorry for two threads in one night, there's no gentle introduction on a maths course!