real analysis - show that all polynomials of odd degree have at least 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.