Problem 4 - A Book of Abstract Algebra

Question

Show that the following is true in \(A[x]\) for any ring \(A:\) For any odd \(n\), (a) \(x+1\) is a factor of \(x^{n}+1\) (b) \(x+1\) is a factor of \(x^{n}+x^{n-1}+\cdots+x+1\)

Step-by-Step Solution

Answer

For odd \(n\), \(x+1\) is a factor of both \(x^n+1\) and \(x^n+x^{n-1}+\cdots+x+1\).

Step 1: Understand the Problem Statement

We need to demonstrate that for any odd integer \( n \), the polynomial \( x+1 \) divides both \( x^n+1 \) and \( x^n+x^{n-1}+\cdots+x+1 \). This means that when these polynomials are divided by \( x+1 \), there should be no remainder.

Step 2: Show that \( x+1 \) divides \( x^n+1 \) for odd \( n \)

To determine if \( x+1 \) is a factor of \( x^n+1 \), we substitute \( x = -1 \) into \( x^n+1 \). We find:\[(-1)^n + 1\]Since \( n \) is odd, \((-1)^n = -1\), making the expression:\[-1 + 1 = 0\]Thus, \( x = -1 \) is a root of \( x^n+1 \), meaning \( x+1 \) is a factor.

Step 3: Show that \( x+1 \) divides \( x^n + x^{n-1} + \cdots + x + 1 \) for odd \( n \)

Consider the polynomial \( f(x) = x^n + x^{n-1} + \cdots + x + 1 \). We substitute \( x = -1 \) into \( f(x) \) to check if \(-1\) is a root:\[(-1)^n + (-1)^{n-1} + \cdots + (-1) + 1\]This simplifies to: \[ n(-1) + 1 \] since the expression has \( n+1 \) terms and will alternate between \(-1\) and \(+1\). Since \( n \) is odd, this results in 0. Hence, \( x = -1 \) is a root of \( f(x) \), implying \( x+1 \) is a factor.

Key Concepts

Polynomial Division

Polynomial division is a method used to divide one polynomial by another, similar to the long division process with numbers. In this exercise, we're focused on dividing polynomials by a binomial like \(x + 1\). The goal is to see if \(x+1\) divides a given polynomial without a remainder. To check this, we typically use substitution or perform the division directly. If substituting a particular value, such as \(x = -1\), into the polynomial results in zero, it indicates that the polynomial is divisible by the binomial, meaning there's no remainder. This is essentially what we're doing when we apply the idea that \(x = -1\) is a root in this problem. The division is confirmed if placing \(x = -1\) leads to zero, affirming \(x + 1\) as a factor.

Odd Integers

Odd integers are numbers that aren't divisible evenly by 2. Examples include \(1, 3, 5, 7,\) and so on. When discussing polynomials, especially in the context of this problem, the parity (whether a number is odd or even) of the exponent \(n\) plays an important role. For any odd \(n\), \((-1)^n\) will equal \(-1\), which is key in calculations involving polynomial roots and factors. This nature of odd numbers—when used as an exponent—helps transform expressions such as \((-1)^n + 1\) into zero, showcasing why an odd number ensures a factorization when a specific substitution is made.

Roots of Polynomials

Roots of a polynomial are the values for which the polynomial equals zero. That is, if \(f(x)\) is a polynomial, then any value \(c\) that makes \(f(c) = 0\) is considered a root. In this exercise, finding that substituting \(x = -1\) into the polynomials results in a zero shows that \(-1\) is a root of both \(x^n + 1\) and \(x^n + x^{n-1} + \cdots + x + 1\). Since \(-1\) is a root, it means \(x + 1\) is a factor of these polynomials. The concept of roots is crucial since identifying roots often guides us to find factors, which simplifies the polynomial and helps solve equations.

Factor Theorem

The Factor Theorem is a fundamental theorem in algebra which states that a polynomial \(f(x)\) has a factor \(x-c\) if and only if \(f(c) = 0\). This is a useful tool for determining factors of a polynomial easily. In this case, by substituting \(x = -1\) into our polynomials and finding the result as zero confirms that \(x + 1\) is indeed a factor. The Factor Theorem helps connect the finding of roots directly to identifying factors, making it an indispensable tool in polynomial algebra. This theorem simplifies the process of factoring polynomials, which otherwise could be quite complex.