91Ó°ÊÓ

An odd-degree polynomial has the general form: f(x) = a_nx^n + a_(n-1)x^(n-1) + ... + a_1x + a_0 where n is an odd number, and a_n ≠ 0. Notice that as x approaches positive or negative infinity, the sign of f(x) will only depend on the leading term, a_nx^n since it will dominate other terms. If a_n > 0, then f(x) will approach positive infinity as x goes to positive infinity, and f(x) will approach negative infinity as x goes to negative infinity. Conversely, if a_n < 0, then f(x) will approach negative infinity as x goes to positive infinity, and f(x) will approach positive infinity as x goes to negative infinity. Step 2: Apply the Intermediate Value Theorem

We will use the Intermediate Value Theorem (IVT) to demonstrate that there is a root in the real numbers for any odd-degree polynomial. Recall that IVT states that if f(x) is continuous and changes sign over an interval [a, b] (i.e., f(a) <= 0 <= f(b) or f(a) >= 0 >= f(b)), then there exists a point c in the interval (a, b) such that f(c) = 0. Let's consider an odd-degree polynomial f(x). We already observed in the previous step that as x goes to positive infinity, f(x) will approach different infinities depending on a_n. Similarly, as x goes to negative infinity, f(x) will also approach different infinities depending on a_n. In each case, there must be a sign change between positive and negative infinity. As f(x) is a polynomial, it is continuous for all real numbers. Therefore, the Intermediate Value Theorem applies, and there exists a point c between positive and negative infinity such that f(c) = 0, which means there is a real root for the polynomial f(x). In conclusion, using the Intermediate Value Theorem, we demonstrated that every polynomial of odd degree over the real numbers has a root in the real numbers.