$\textbf{Exercise 9.}$ Prove that every polynomial of odd degree with real coefficients has a real zero. $\textbf{Solution 9.}$ Suppose $p$ is a polynomial of odd degree and real coefficients. By 4.16, $p$ is a constant times the product of factors of the form $x - \lambda$ and/or $x^2 + bx + c$, where $\lambda, b, c \in \mathbb{R}$. Not all the factors can be of the form $x^2 + bx + c$, because otherwise $p$ would have even degree. Thus at least one factor is of the form $x - \lambda$. Any such $\lambda$ is a real zero of $p$. $\textit{Comment:}$ Here is another proof, using calculus but not using 4.16. Suppose $p$ is a polynomial of odd degree $m$. We can write $p$ in the form $\begin{equation*} p(x) = a_0 + a_1x + \cdots + a_mx^m, \end{equation*}$ where $a_0, \ldots, a_m \in \mathbb{R}$ and $a_m \neq 0$. Replacing $p$ with $-p$ if necessary, we can assume that $a_m > 0$. Now $\begin{equation*} p(x) = x^m \left(\frac{a_0}{x^m} + \frac{a_1}{x^{m-1}} + \cdots + \frac{a_{m-1}}{x} + a_m\right). \end{equation*}$ This implies that $\begin{equation*} \lim_{x \to -\infty} p(x) = -\infty \text{ and } \lim_{x \to \infty} p(x) = \infty. \end{equation*}$ The intermediate value theorem now implies that there is a real number $\lambda$ such that $p(\lambda) = 0$. In other words, $p$ has a real zero. ---  --- Commentary: $\textbf{Exercise 9.}$ Prove that every polynomial of odd degree with real coefficients has a real zero. $\textbf{Solution 9.}$ Suppose $p$ is a polynomial of odd degree and real coefficients. By 4.16, $p$ is a constant times the product of factors of the form $x - \lambda$ and/or $x^2 + bx + c$, where $\lambda, b, c \in \mathbb{R}$. Not all the factors can be of the form $x^2 + bx + c$, because otherwise $p$ would have even degree. Thus at least one factor is of the form $x - \lambda$. Any such $\lambda$ is a real zero of $p$. $\textit{Comment:}$ Here is another proof, using calculus but not using 4.16. Suppose $p$ is a polynomial of odd degree $m$. We can write $p$ in the form $\begin{equation*} p(x) = a_0 + a_1x + \cdots + a_mx^m, \end{equation*}$ where $a_0, \ldots, a_m \in \mathbb{R}$ and $a_m \neq 0$. Replacing $p$ with $-p$ if necessary, we can assume that $a_m > 0$. Now $\begin{equation*} p(x) = x^m \left(\frac{a_0}{x^m} + \frac{a_1}{x^{m-1}} + \cdots + \frac{a_{m-1}}{x} + a_m\right). \end{equation*}$ This implies that $\begin{equation*} \lim_{x \to -\infty} p(x) = -\infty \text{ and } \lim_{x \to \infty} p(x) = \infty. \end{equation*}$ The intermediate value theorem now implies that there is a real number $\lambda$ such that $p(\lambda) = 0$. In other words, $p$ has a real zero. $\textit{Commentary:}$ This exercise is a classic result in algebra and real analysis. It says that a real polynomial of odd degree must have at least one real root. The proof given in the solution uses the factorization theorem for real polynomials (4.16), which says that any real polynomial can be factored into linear and irreducible quadratic factors. If the polynomial has odd degree, at least one of these factors must be linear, giving a real root. The alternative proof uses calculus: it shows that a real polynomial of odd degree with positive leading coefficient goes to $-\infty$ at $-\infty$ and $\infty$ at $\infty$, so it must cross the $x$-axis somewhere by the intermediate value theorem. This result has numerous applications in algebra, geometry, and analysis. $\textit{Example:}$ Consider the polynomial $p(x) = x^3 - x + 1$. This has odd degree (3) and real coefficients. By the theorem, it must have a real root. Indeed, we can factor it as $\begin{equation*} p(x) = (x - 1)(x^2 + x - 1), \end{equation*}$ so $x = 1$ is a real root. (The other roots are complex: $x = \frac{-1 \pm \sqrt{5}}{2}$.) Alternatively, we see that $\lim_{x \to -\infty} p(x) = -\infty$ and $\lim_{x \to \infty} p(x) = \infty$, so $p$ must cross the $x$-axis at some point, which is a real root.