Chapter 26: Problem 2
Let \(F\) be a field. Prove that each of the following is true in \(F[x] .\) \((x-c) \mid(p(x)-p(c))\)
Short Answer
Expert verified
The polynomial \((x-c)\) divides \(p(x)-p(c)\) in \(F[x]\) because \(p(x) - p(c) = (x-c)q(x)\) for some polynomial \(q(x)\).
Step by step solution
01
Understand the Problem
You are given a polynomial \( p(x) \) in the polynomial ring \( F[x] \), where \( F \) is a field. To prove \((x-c) \mid(p(x)-p(c))\), you need to show that \( x-c \) is a divisor of \( p(x) - p(c) \) in \( F[x] \). This means there exists a polynomial \( q(x) \) in \( F[x] \) such that \( p(x) - p(c) = (x-c) \cdot q(x) \).
02
Apply Remainder Theorem
Recall the Remainder Theorem for polynomials: For any polynomial \( p(x) \) and any scalar \( c \), there exists a unique polynomial \( q(x) \) and a remainder \( r \) such that \( p(x) = (x-c)q(x) + r \), where \( r \) is a constant (a scalar from \( F \)). Moreover, the theorem tells us that \( r = p(c) \).
03
Substitute and Simplify
Write \( p(x) - p(c) \) as \( (x-c)q(x) + p(c) - p(c) \). This simplifies to \( (x-c)q(x) \), which shows \( p(x) - p(c) = (x-c)q(x) \).
04
Conclude the Proof
Since \( p(x) - p(c) \) can be expressed as \( (x-c)q(x) \) where \( q(x) \) is a polynomial in \( F[x] \), it follows that \( x-c \) divides \( p(x) - p(c) \) in the polynomial ring \( F[x] \). Hence, we have proved the statement.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Division
Polynomial division is a technique very similar to long division with numbers, but it is done with polynomials instead. When you divide a polynomial by another polynomial, you aim to find a quotient and a remainder. The idea is that for any polynomials \( f(x) \) and \( g(x) \), where \( g(x) eq 0 \), you can express \( f(x) \) as:
This process is crucial in determining how many times one polynomial can "fit" into another, while factoring any left-over terms into a smaller polynomial.
In the context of our exercise, using polynomial division helps illustrate how \((x-c)\) is a divisor of \(p(x) - p(c)\). The key is setting up the division so that the remainder directly connects to the value of the polynomial at a specific point \( c \).
- \( f(x) = q(x)g(x) + r(x) \)
This process is crucial in determining how many times one polynomial can "fit" into another, while factoring any left-over terms into a smaller polynomial.
In the context of our exercise, using polynomial division helps illustrate how \((x-c)\) is a divisor of \(p(x) - p(c)\). The key is setting up the division so that the remainder directly connects to the value of the polynomial at a specific point \( c \).
Field Theory
Field Theory is a branch of mathematics that studies fields. A field is a set equipped with two operations: addition and multiplication. It has structures similar to the real number system, allowing us to do arithmetic without leaving the field.
In this problem, we deal with a polynomial ring \( F[x] \), where \( F \) is a field.
Since a field has no divisors of zero, every nonzero element has a multiplicative inverse. This helps ensure that polynomials can behave in a controllable way when working in \( F[x] \).
In this problem, we deal with a polynomial ring \( F[x] \), where \( F \) is a field.
- Fields are fundamental in polynomial operations since they ensure that division (except by zero) is always possible, which is crucial for finding quotients and remainders in polynomial division.
Since a field has no divisors of zero, every nonzero element has a multiplicative inverse. This helps ensure that polynomials can behave in a controllable way when working in \( F[x] \).
Remainder Theorem
The Remainder Theorem is a straightforward yet powerful tool in polynomial algebra. It states that for a polynomial \( p(x) \), if you divide it by \( x-c \), the remainder of this division is simply \( p(c) \).
This theorem is fundamental for quickly finding the value of a polynomial at a particular point because:
In the context of our exercise, the Remainder Theorem allows us to express \( p(x)-p(c) = (x-c)q(x) \). The remainder in this division disappears, illustrating \((x-c)\) as a true divisor.
This theorem is fundamental for quickly finding the value of a polynomial at a particular point because:
- If \( p(x) = (x-c)q(x) + r \) and the remainder \( r \) must be a constant, then \( p(c) = r \).
In the context of our exercise, the Remainder Theorem allows us to express \( p(x)-p(c) = (x-c)q(x) \). The remainder in this division disappears, illustrating \((x-c)\) as a true divisor.
Polynomial Ring
A polynomial ring is a mathematical construct that consists of all polynomials with coefficients from a given field \( F \). When we speak of a ring, we refer to a set that allows both addition and multiplication operations and follows certain algebraic rules.
In the context of our exercise, we consider the polynomial ring \( F[x] \), where
Why does \( F[x] \) matter? It is because it provides a structured framework to explore polynomial equations and perform division, leading us to results like the division of \( p(x) - p(c) \) by \( x-c \). This realm ensures that operations such as factoring, finding divisors, and solving equations remain algebraically consistent.
In the context of our exercise, we consider the polynomial ring \( F[x] \), where
- \( F \) is a field (ensuring well-behaved arithmetic).
- \( x \) is an indeterminate (a placeholder for variables in our polynomials).
Why does \( F[x] \) matter? It is because it provides a structured framework to explore polynomial equations and perform division, leading us to results like the division of \( p(x) - p(c) \) by \( x-c \). This realm ensures that operations such as factoring, finding divisors, and solving equations remain algebraically consistent.