It is proved that the polynomial A0 * x ^ n + A1 * x ^ n-1 + A2 * x ^ n-2 +. +... An = 0 has at least one real root when n is odd. (A0! = 0)

It is proved that the polynomial A0 * x ^ n + A1 * x ^ n-1 + A2 * x ^ n-2 +. +... An = 0 has at least one real root when n is odd. (A0! = 0)

Let A0 > 0
We prove that when x is a sufficiently large positive real number, the polynomial takes a positive value, and when x is a sufficiently large negative real number, it takes a negative value
So there is a zero point, the real root
In fact, when | x | > | A1 / A0 | + | A2 / A0 | +... + | an / A0 | + 1
There are | A0 · x ^ n | > | A1 · x ^ n | + | A2 · x ^ n | +... + | an · x ^ n|
> |a1·x^(n-1)|+|a2·x^(n-2)|+...+|an|
≥ |a1·x^(n-1)+a2·x^(n-2)+...+an|.
From A0 > 0, if x > 0, then A0 · x ^ n > 0, there is A0 · x ^ n + A1 · x ^ (n-1) + A2 · x ^ (n-2) +... + an > 0
If x < 0, then A0 · x ^ n < 0, there are A0 · x ^ n + A1 · x ^ (n-1) + A2 · x ^ (n-2) +... + an < 0
A 0 · x ^ n + a 1 · x ^ (n-1) + a 2 · x ^ (n-2) +... + an is continuous with respect to x, so there is a zero
Another method, by the basic theorem of algebra, the equation of degree n has n complex roots
However, the imaginary roots of polynomial equation with real coefficients are paired, but n is odd, so there are real roots