Implications of the Condition

Oct 13, 2025 by wbrum.org 0 views

Does $q(x)=q(2x)$ for all $x$ imply that $q(x)$ is a constant polynomial?

In algebra and precalculus, polynomials are a fundamental concept that plays a crucial role in solving various mathematical problems. A polynomial is an expression consisting of variables and coefficients combined using only addition, subtraction, and multiplication, and non-negative integer exponents. In this article, we will explore the relationship between a polynomial $q(x)$ and its behavior when evaluated at $2x$ . Specifically, we will investigate whether the condition $q(x) = q(2x)$ for all $x$ implies that $q(x)$ is a constant polynomial.

The problem we are trying to solve is related to a given polynomial $P(x)$ with real coefficients, which satisfies the equation $(x-8)P(2x) = 8(x-1)P(x)$ . By solving this equation, we find that $P(x) = (x-8)(x-4)(x-2)q(x)$ , where $q(x)$ is a polynomial. Our goal is to determine whether the condition $q(x) = q(2x)$ for all $x$ implies that $q(x)$ is a constant polynomial.

Analyzing the Condition $q(x) = q(2x)$

To understand the implications of the condition $q(x) = q(2x)$ , let's analyze its behavior. Suppose $q(x)$ is a polynomial of degree $n$ , then we can write $q(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$ , where $a_n \neq 0$ . Now, let's evaluate $q(2x)$ :

q(2x) = a_n (2x)^n + a_{n-1} (2x)^{n-1} + \ldots + a_1 (2x) + a_0

Simplifying the expression, we get:

q(2x) = 2^n a_n x^n + 2^{n-1} a_{n-1} x^{n-1} + \ldots + 2a_1 x + a_0

Comparing the coefficients of $q(x)$ and $q(2x)$ , we can see that the coefficients of $q(2x)$ are all multiples of the corresponding coefficients of $q(x)$ , except for the constant term, which remains the same.

Implications of the Condition

Now that we have analyzed the behavior of $q(x)$ and $q(2x)$ , let's consider the implications of the condition $q(x) = q(2x)$ for all $x$ . If $q(x)$ is a polynomial of degree $n$ , then the condition $q(x) = q(2x)$ implies that the coefficients of $q(x)$ must be such that the coefficients of $q(2x)$ are all multiples of the corresponding coefficients of $q(x)$ . This means that the coefficients of $q(x)$ must be zero, except for the constant term, which must be equal to the constant term of $q(2x)$ .

In conclusion, the condition $q(x) = q(2x)$ for all $x$ implies that $q(x)$ is a constant polynomial. This is because the coefficients of $q(x)$ must be zero, except for the constant term, which must be equal to the constant term of $q(2x)$ . Therefore, we can conclude that $q(x)$ is a constant polynomial.

To illustrate this result, let's consider an example. Suppose we have a polynomial $q(x) = x^2 + 2x + 1$ . Evaluating $q(2x)$ , we get:

q(2x) = (2x)^2 + 2(2x) + 1 = 4x^2 + 4x + 1

Comparing the coefficients of $q(x)$ and $q(2x)$ , we can see that the coefficients of $q(2x)$ are not all multiples of the corresponding coefficients of $q(x)$ . Therefore, the condition $q(x) = q(2x)$ is not satisfied, and we cannot conclude that $q(x)$ is a constant polynomial.

To further illustrate the result, let's consider a counterexample. Suppose we have a polynomial $q(x) = x^2 + 2x + 1$ . Evaluating $q(2x)$ , we get:

q(2x) = (2x)^2 + 2(2x) + 1 = 4x^2 + 4x + 1

In this article, we have explored the relationship between a polynomial $q(x)$ and its behavior when evaluated at $2x$ . We have shown that the condition $q(x) = q(2x)$ for all $x$ implies that $q(x)$ is a constant polynomial. This result has important implications for the study of polynomials and their properties. We hope that this article has provided a clear and concise explanation of this result, and we look forward to hearing your thoughts and feedback.
Q&A: Does $q(x)=q(2x)$ for all $x$ imply that $q(x)$ is a constant polynomial?

In our previous article, we explored the relationship between a polynomial $q(x)$ and its behavior when evaluated at $2x$ . We showed that the condition $q(x) = q(2x)$ for all $x$ implies that $q(x)$ is a constant polynomial. In this article, we will answer some frequently asked questions related to this result.

Q: What is the significance of the condition $q(x) = q(2x)$ ?

A: The condition $q(x) = q(2x)$ is significant because it implies that the polynomial $q(x)$ is a constant polynomial. This is a fundamental result in algebra and precalculus, and it has important implications for the study of polynomials and their properties.

Q: How do we prove that $q(x) = q(2x)$ implies that $q(x)$ is a constant polynomial?

A: To prove that $q(x) = q(2x)$ implies that $q(x)$ is a constant polynomial, we need to analyze the behavior of $q(x)$ and $q(2x)$ . We can do this by comparing the coefficients of $q(x)$ and $q(2x)$ . If the coefficients of $q(2x)$ are all multiples of the corresponding coefficients of $q(x)$ , except for the constant term, then we can conclude that $q(x)$ is a constant polynomial.

Q: What is the relationship between the coefficients of $q(x)$ and $q(2x)$ ?

A: The coefficients of $q(2x)$ are all multiples of the corresponding coefficients of $q(x)$ , except for the constant term. This means that the coefficients of $q(x)$ must be zero, except for the constant term, which must be equal to the constant term of $q(2x)$ .

Q: Can we have a polynomial $q(x)$ that satisfies the condition $q(x) = q(2x)$ but is not a constant polynomial?

A: No, we cannot have a polynomial $q(x)$ that satisfies the condition $q(x) = q(2x)$ but is not a constant polynomial. If $q(x)$ satisfies the condition $q(x) = q(2x)$ , then it must be a constant polynomial.

Q: What are some examples of polynomials that satisfy the condition $q(x) = q(2x)$ ?

A: Some examples of polynomials that satisfy the condition $q(x) = q(2x)$ are:

$q(x) = 1$
$q(x) = x$
$q(x) = x^2 + 2x + 1$

These polynomials are all constant polynomials, and they satisfy the condition $q(x) = q(2x)$ .

Q: What are some examples of polynomials that do not satisfy the condition $q(x) = q(2x)$ ?

A: Some examples of polynomials that do not satisfy the condition $q(x) = q(2x)$ are:

$q(x) = x^2 + 2x$
$q(x) = x^3 + 2x^2 + 3x + 1$

These polynomials do not satisfy the condition $q(x) = q(2x)$ , and they are not constant polynomials.

In this article, we have answered some frequently asked questions related to the result that $q(x) = q(2x)$ implies that $q(x)$ is a constant polynomial. We hope that this article has provided a clear and concise explanation of this result, and we look forward to hearing your thoughts and feedback.