Prove that every polynomial of odd degree with real coefficients has at least one real root.
Get the complete, step-by-step math solution for: "Prove that every polynomial of odd degree with real coefficients has at least one real root.". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Define the polynomial
Let P(x) be a polynomial of odd degree n, where n is a positive odd integer. The coefficients an,an−1,…,a0 are real numbers, and an=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 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, anxn, dominates the polynomial's behavior. Since n is odd, xn approaches infinity as x approaches infinity. Therefore, the sign of P(x) as x→∞ is determined by the sign of an. If an>0, then P(x)→∞. If an<0, then P(x)→−∞.
Step 3: Analyze the limit as x→−∞
Next, we examine the behavior of the polynomial for very large negative values of x. As x approaches negative infinity, the term xn (where n is odd) approaches negative infinity. Therefore, the sign of P(x) as x→−∞ will be opposite to the sign of an. If an>0, then P(x)→−∞. If an<0, then P(x)→∞.
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 an>0, then P(x) goes from −∞ to ∞. If an<0, then P(x) goes from ∞ to −∞. 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.