This is the same idea as Jhevon's solution to the first problem.
Given y(x) = ax^n + ... + b, where a is not 0 and n is odd, note that
lim[y(x), x --> -(infinity)] --> (-/+)(infinity) <-- Choice of sign depends on the sign of the coefficient a.
lim[y(x), x--> (infinity)] --> (+/-)(infinity)
Since the polynomial function either goes from -(infinity) to (infinity) or from (infinity) to -(infinity) we know it has to cross the x-axis at least once.
-Dan
Consider the odd degree polynomial, P(x) = ax^n + bx^(n-1) + cx^(n-2) + ..., where n is odd.
Assume a>0
Now P(k) = ak^n + bk^(n-1) + ck^(n-2) + ...
=> P(k) = ak^n(1 + [bk^(n-1)]/ak^n + [ck^(n-2)]/ak^n + ...)
=> P(k) = ak^n(1 + b/ak + c/ak^2 + ...)
=> lim{k--> -inf}P(k) = lim{k--> -inf}ak^n(1 + b/ak + c/ak^2 + ...) = lim{k--> -inf}aK^n
since n is odd,
=> lim{k--> -inf}ak^n = -inf
So there is some k for which P(k) < 0
Now P(l) = al^n + bl^(n-1) + cl^(n-2) + ...
=> P(l) = al^n(1 + [bl^(n-1)]/al^n + [cl^(n-2)]/al^n + ...)
=> P(l) = al^n(1 + b/al + c/al^2 + ...)
=> lim{l--> inf}P(l) = lim{l--> inf}al^n(1 + b/al + c/al^2 + ...) = lim{l--> inf}al^n
since n is odd,
=> lim{l--> inf}al^n = inf
So there is some l for which P(l) > 0
Now, the opposte happens for a < 0, that is, P(k)>0 and P(l)<0, either way, 0 is between P(k) and P(l). Thus by the intermediate value theorem, there is some c between k and l such that P(c) = 0. So for x = c, P(x) has a real root
EDIT: Ah man, Dan beat me to it. I went a lot slower for the second one than he did though, I guess you can use his way