Prove that every polynomial of odd degree with real coefficients has at least one real root.

Step 1: Define the polynomial

Let $P(x)$ be a polynomial of odd degree $n$, where $n$ is a positive odd integer. The coefficients $a_n, a_{n-1}, \dots, a_0$ are real numbers, and $a_n \neq 0$. We need to show that there exists at least one real number $c$ such that $P(c) = 0$.

Step 2: Analyze the limit as $x \to \infty$

To understand the behavior of the polynomial for very large positive values of $x$, we examine the limit as $x$ approaches infinity. For large $x$, the term with the highest power, $a_n x^n$, dominates the polynomial's behavior. Since $n$ is odd, $x^n$ approaches infinity as $x$ approaches infinity. Therefore, the sign of $P(x)$ as $x \to \infty$ is determined by the sign of $a_n$. If $a_n > 0$, then $P(x) \to \infty$. If $a_n < 0$, then $P(x) \to -\infty$.

Step 3: Analyze the limit as $x \to -\infty$

Next, we examine the behavior of the polynomial for very large negative values of $x$. As $x$ approaches negative infinity, the term $x^n$ (where $n$ is odd) approaches negative infinity. Therefore, the sign of $P(x)$ as $x \to -\infty$ will be opposite to the sign of $a_n$. If $a_n > 0$, then $P(x) \to -\infty$. If $a_n < 0$, then $P(x) \to \infty$.

Step 4: Apply the Intermediate Value Theorem

From the limits, we see that $P(x)$ takes on both very large positive and very large negative values. Specifically, if $a_n > 0$, then $P(x)$ goes from $-\infty$ to $\infty$. If $a_n < 0$, then $P(x)$ goes from $\infty$ to $-\infty$. Since $P(x)$ is a polynomial, it is continuous everywhere. By the Intermediate Value Theorem, if a continuous function takes on two values, it must take on every value in between. Since $P(x)$ takes on both positive and negative values, it must cross the $x$ -axis at least once, meaning there exists at least one real root $c$ such that $P(c) = 0$.

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