Question

Prove that any polynomial of odd degree has atleast one real root

Answer 1

Solution:

We want to show that if $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ is a polynomial with n odd and $a_{\mathrm{n}} \neq 0$ , then there is a real number c, such that P(c)=0

Firstly we know that every polynomial is continuous on the real line. We also know that

$\lim _{n \rightarrow \infty} \frac{P(x)}{a_{n} x^{n}}=1 \text { and } \lim _{n \rightarrow-\infty} \frac{P(x)}{a_{n} x^{n}}=1$

Consequently for $|x|$ large enough, $P(x)$ and $a_nx^n$ have the same sign. But $a_nx^n$ has opposite signs for positive $x$ and negative $x$. Thus it follows that if $a_n>0$, there are real numbers $x_0<x_1$ such that $P(x_0)<0$ and $P(x_1)>0$. Similarly, if $a_n<0$, we can find $x_0<x_1$ such that $P(x_0)>0$ and $P(x_1)<0$. In each case, it now follows directly from the Intermediate Value Theorem that (for d = 0) there is a real number $c\in\left[x_{0}, x_{1}\right]\ with\ P(c)=0$