Show that any polynomial of odd degree has at least one real root.
Another good method is to note the fact that in a degree n polynomial has n roots (counting with multiplicity) as is algebraicly closed . But for every complex number, it's conjugate is also a root, you can check this easily enough. But since there are an odd number of roots that means one of the conjugates must be itself, but this means one of the roots is purely real.
This is a more algebraic argument of course, but it works all the same.