show that all polynomials of odd degree have atleast one real root?
What you want to show is this:
If n is odd, then any equation
x^n + a_(n - 1)x^(n - 1) + ... + a_0 = 0
has a root.
Proof:
We obviously want to consider the function
f(x) = x^n + a_(n - 1)x^(n -1) + ... a_0;
we would like to prove that f is sometimes positive and sometimes negative. The intuitive idea is that for large |x|, the function is very much like g(x) = x^n and, since n is odd, this function is positive for large positive x and negative for large negative x. A little algebra is all we need to make this intuitive idea work.
The proper analysis of the function f depends on writing
f(x) = x^n + a_(n - 1)x^(n -1) + ... a_0 = x^n(1 + a_(n-1)/x + ... + a_0/x^n).
Note that
|a_(n - 1)/x + a_(n -2)/x^2 + ... + a_0/x^n| ≤ |a_(n-1)|/|x| + ... + |a_0|/|x^n|
Consequently, if we choose x satisfying
(*) |x| > 1, 2n|a_(n-1)|, ..., 2n|a_0|,
then |x^k| > |x| and
|a_(n - k)|/|x^k| < |a_(n - k)|/|x| < |a_(n-k)|/2n|a_(n-k)| = 1/2n,
so
|a_(n-1)/x + a_(n-2)/x^2 + ... + a_0/x^n| ≤ 1/2n + ... + 1/2n = 1/2
In other words,
-1/2 ≤ a_(n-1)/x + ... + a_0/x^n ≤ 1/2,
which implies that
1/2 ≤ 1 + a_(n-1)/x + ... + a_0/x^n.
Therefore, if we choose an x_1 > 0 which satisfies (*), then
[(x_1)^n]/2 ≤ (x_1)^n(1 + a_(n-1)/x_1 + ... + a_0/(x_1)^n) = f(x_1)
so that f(x_1) > 0. On the other hand, if x_2 < 0 satisfies (*), then (x_2)^n < 0 and
[(x_2)^n]/2 ≥ (x_2)^n(1 + a_(n-1)/x_2 + ... + a_0/(x_2)^n) = f(x_2),
so that f(x_2) < 0.
Now applying the Intermediate Value theorem to the interval [x_1, x_2] we conclude that there is an x in [x_1, x_2] such that f(x) = 0.