Problem 15 - Algebra Form and Function

Question

Explain why \(p(x)=x^{5}+3 x^{3}+2\) must have at least one zero.

Step-by-Step Solution

Answer

Question: Prove that the polynomial \(p(x)=x^5 + 3x^3 + 2\) has at least one zero. Answer: By using the Intermediate Value Theorem and evaluating the polynomial at \(x=-1\) and \(x=1\), we found that \(p(-1) < 0\) and \(p(1) > 0\). Since the polynomial is continuous on its entire domain, there must exist a value \(c\) between \(-1\) and \(1\) such that \(p(c) = 0\). Thus, the polynomial \(p(x)=x^5 + 3x^3 + 2\) has at least one zero.

Step 1: Identify the function and its range

We are given the polynomial function \(p(x)=x^5 + 3x^3 + 2\). Since polynomials are continuous on their entire domain \((-\infty, \infty)\), we can apply the IVT.

Step 2: Find the values of \(p(x)\) at some points

Our goal is to find two points \(a\) and \(b\) such that \(p(a)\) and \(p(b)\) have opposite signs. Let's evaluate \(p(x)\) at \(x=-1\) and \(x = 1\): \(p(-1) = (-1)^5 + 3(-1)^3 +2=-1-3+2 =-2 < 0\) \(p(1) = (1)^5 + 3(1)^3 + 2=1+3+2 = 6 > 0\)

Step 3: Apply the Intermediate Value Theorem

Since \(p(-1) < 0\) and \(p(1) > 0\), we can apply the Intermediate Value Theorem. Since \(p(x)\) is continuous on the entire domain, there must exist a value \(c\) between \(-1\) and \(1\), such that \(p(c) = 0\). Therefore, the polynomial \(p(x)=x^5 + 3x^3 + 2\) has at least one zero.

Key Concepts

Polynomial Function

A polynomial function is an expression consisting of variables and coefficients, composed together using addition, subtraction, and multiplication. These functions are built on terms that are powers of a variable, such as

• Linear polynomials like \[ ax + b \]
• Quadratic polynomials like \[ ax^2 + bx + c \]
• Higher degree polynomials like \[ x^5 + 3x^3 + 2 \], as in our case

In each term, the exponent of the variable is always a whole number. Polynomials are key building blocks in algebra, enabling the solving of equations, modeling of relationships, and description of various mathematical phenomena. Polynomial functions can represent a wide range of behaviors and are pivotal in calculus when analysing changes and trends in data.
Moreover, the highest power of the variable is known as the degree of the polynomial. In \( p(x)=x^5+3x^3+2 \), the degree is 5. This indicates that the graph of the polynomial can have up to 5 real roots or zero points, depending on the specific configuration.

Continuous Function

Continuous functions are a fundamental concept in calculus. A function is continuous over its domain if you can draw its graph without lifting your pencil. Simply put, there are no breaks, jumps, or holes. For any small change in the input, there will be only a very small change in the output.
Polynomials are an example of continuous functions. This is because they are defined for all real numbers and have no disruptions in their graphs. This continuity makes them behave predictably over their entire domain,

• Ensuring that we can apply useful mathematical theorems, like the Intermediate Value Theorem
• Allowing consistent evaluation of limits
• Making differentiation and integration straightforward

For the polynomial \( p(x) = x^5 + 3x^3 + 2 \), its continuity across all real numbers means it provides a reliable and consistent behavior in this range, crucial for finding its zeros.

Zero of a Polynomial

The zero of a polynomial is an input value that makes the entire polynomial equal to zero. It is where the graph of the polynomial touches or crosses the x-axis. Any polynomial with degree \( n \) can have up to \( n \) zeros, though not all might be real or distinct.
Finding these zeros is crucial because they often represent solutions to equations, and in some contexts, they can indicate significant points such as equilibrium in physical systems or break-even points in economics.
In our exercise, to show that the polynomial \( p(x)=x^5 + 3x^3 + 2 \) has at least one zero, the Intermediate Value Theorem (IVT) is utilized. This theorem states that for any continuous function that has values of opposite signs at two points, there's at least one root in the interval between them.

• Evaluating the function at \( x = -1 \) and \( x = 1 \) gives opposite signs
• Letting us affirm the presence of a zero between these points by IVT

Thus, the IVT simplifies the process of finding, or at least confirming the existence of, the zero of the polynomial without explicitly calculating it.