Problem 52 - Algebra for College Students

Question

Verify that \(x-1\) is a factor of \(x^{n}-1\) for all positive integral values of \(n\). See below

Step-by-Step Solution

Answer

Yes, \(x-1\) is a factor of \(x^n - 1\) for all positive integers \(n\).

Step 1: Understand the Problem

Our goal is to verify that \(x-1\) is a factor of \(x^n - 1\) for all positive integer values of \(n\). This means we need to show that evaluating \(x^n - 1\) at \(x = 1\) gives us zero, or alternatively, that \(x^n - 1\) can be expressed as the product of \(x-1\) and another polynomial for all \(n\).

Step 2: Apply the Factor Theorem

The Factor Theorem states that \(x-a\) is a factor of a polynomial \(P(x)\) if \(P(a) = 0\). Let's apply this theorem to the polynomial \(P(x) = x^n - 1\) with \(a = 1\).

Step 3: Compute \(P(1)\)

Substitute \(x = 1\) into \(P(x) = x^n - 1\). This gives:\[P(1) = 1^n - 1 = 1 - 1 = 0\].Since \(P(1) = 0\), the Factor Theorem confirms that \(x - 1\) is a factor of \(x^n - 1\).

Step 4: Discuss General Factorization Approach

We can consider \(x^n - 1\) as a difference of powers. The formula for factoring a difference of powers is given by:\[ x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + x + 1) \].This expression directly shows that \(x-1\) is a factor of \(x^n - 1\). The second factor is a polynomial of degree \(n-1\), confirming the factorization.

Key Concepts

Polynomial Division

Polynomial division is a powerful tool in algebra that allows us to simplify expressions and solve polynomial equations. In its essence, polynomial division resembles long division, which you may be familiar with from arithmetic. When you divide one polynomial by another, you can think of it like arranging terms, and systematically subtracting them, until you obtain the remainder.

For instance, when dividing the polynomial \(x^n - 1\) by \(x - 1\), we use this method to determine the quotient, which will be a polynomial of degree \(n - 1\). This quotient is actually the polynomial \(x^{n-1} + x^{n-2} + \cdots + x + 1\), which becomes evident through polynomial division. Here are some key aspects to keep in mind during polynomial division:

• Align the terms in descending order of their degrees.
• Divide the leading term of the dividend by the leading term of the divisor.
• Subtract the resulting polynomial from the original polynomial.
• Repeat the process with the new polynomial obtained.

This systematic approach either gives you the quotient and remainder or verifies that one polynomial is a factor of another, like \(x - 1\) being a factor of \(x^n - 1\) without a remainder.

Difference of Powers

The expression \(x^n - 1\) is known in algebra as a difference of powers. This term is applicable when two quantities with the same base are subtracted from each other, leading to a factorable expression.

When we explore \(x^n - 1\), it can be broken down using a specific formula: \[ x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + x + 1) \]This formula is derived from the fact that when you subtract 1 from \(x^n\), you effectively subtract \(x\) raised to the power zero from \(x\) raised to the power \(n\). This results in a series of terms that decrease in power from \(n-1\) to zero.

Understanding this pattern is crucial because it allows you to swiftly recognize the factorization and validate that \(x - 1\) is indeed a factor. This method is pivotal when solving polynomial equations and aids in understanding the behavior of polynomial graphs, specifically where they cross the x-axis.

Integral Roots

In algebra, integral roots are solutions to polynomial equations that are integers. These roots satisfy the equation when substituted in place of the variable.

For the given polynomial \(x^n - 1\), testing \(x = 1\) gives us \(1^n - 1 = 0\). This calculation means that \(x = 1\) is an integral root, showing \(x - 1\) is a factor. Each root corresponds to a factor, and finding integral roots often simplifies polynomial factorization.

• The Factor Theorem tells us that if \(P(a) = 0\), then \(x-a\) is a factor of the polynomial \(P(x)\).
• For every factor of the form \(x-a\), \(a\) is an integral root if it's an integer.

Integral roots are significant in simplifying and solving polynomial equations, and recognizing them helps in the construction and decomposition of polynomials into smaller, manageable parts.