Problem 53 - Algebra for College Students

Question

(a) Verify that \(x-y\) is a factor of \(x^{n}-y^{n}\) for all positive integral values of \(n\). See below (b) Verify that \(x+y\) is a factor of \(x^{n}-y^{n}\) for all even positive integral values of \(n\). See below (c) Verify that \(x+y\) is a factor of \(x^{n}+y^{n}\) for all odd positive integral values of \(n\). See below

Step-by-Step Solution

Answer

(a) Yes, \(x-y\) is a factor for all \(n\). (b) Yes, \(x+y\) is a factor for even \(n\). (c) Yes, \(x+y\) is a factor for odd \(n\).

Step 1: Use Polynomial Division

To show that \(x - y\) is a factor of \(x^n - y^n\), we first express \(x^n - y^n\) as a difference of powers. Using polynomial division, we divide \(x^n - y^n\) by \(x - y\) and look for a remainder. By the Factor Theorem, if \(x - y\) divides \(x^n - y^n\) with a remainder of zero, then \(x - y\) is a factor of \(x^n - y^n\).

Step 2: Apply the Factor Theorem

When dividing \(x^n - y^n\) by \(x - y\), substitute \(x = y\). Substitute values into the derived quotient to check if the remainder is 0. The remainder term vanishes, confirming that \(x - y\) is indeed a factor when \(x = y\). Repetition of this step for other powers and positive integers \(n\) confirms the factorization.

Step 3: Use the Even Power Expansion

For \(x + y\) as a factor of \(x^n - y^n\) with even \(n\), expand \((x^2)^k - (y^2)^k\), where \(n = 2k\), using the formula \((a^k - b^k)(a^k + b^k)\). Since \(n\) is even, substitute \(x = -y\) to see if the expression simplifies to zero. This step shows that the factorization results in zero, confirming \(x + y\) is a factor.

Step 4: Use the Odd Power Expansion

For \(x + y\) as a factor of \(x^n + y^n\) with odd \(n\), write \(x^n + y^n\) in terms of \((c + d)^k\), where \(c=x\) and \(d=y\), showing even powers sum under subtraction. Substituting \(x = -y\) in the equation demonstrates simplification to zero, implying that the entire expression becomes zero for odd \(n\). Thus, \(x + y\) is a factor when \(n\) is odd.

Step 5: Combine Factorization Results

Combine all insights from the expansions, substitutions, and results from previous steps to conclude the factorization confirmed in each case. Verify with multiple values to ensure these observations hold true, supporting sound mathematical reasoning.

Key Concepts

Factor Theorem

The Factor Theorem is an essential principle in algebra, especially when dealing with polynomial expressions. It states that a polynomial \( f(x) \) has a factor \((x - c)\) if and only if \( f(c) = 0 \). This means that your polynomial becomes zero when you substitute \( x \) with \( c \).

For instance, if you need to check if \( x - y \) is a factor of a polynomial like \( x^n - y^n \), simply substitute \( x = y \) into the polynomial. If the polynomial evaluates to zero, then, according to the Factor Theorem, \( x - y \) is indeed a factor.

Applying this theorem helps simplify verifying whether simple expressions are factors of more complex polynomials, and it plays a crucial role in polynomial division by helping in identifying divisors.

Even and Odd Powers

Understanding even and odd powers is fundamental when working with polynomials, particularly those involving variables raised to such powers.

**Even Powers**: When \( n \) is even, a polynomial expression like \( x^n \) will have certain properties. For example, \((x+y)\) is a factor of \(x^n - y^n\) provided \(n\) is even. This is essential knowledge when performing polynomial division, as it ensures that certain symmetrical properties hold true, allowing simplification.

**Odd Powers**: With odd power polynomials such as \(x^n + y^n\) where \(n\) is odd, different factorization properties emerge. For these expressions, \(x+y\) is a factor. Using these properties can help reveal deep insights into the behavior of polynomials and how they can be simplified or factored further.

Recognizing the role of even and odd powers enhances your ability to predict the factorizability of polynomial expressions and simplify them accordingly.

Factorization

The process of factorization involves breaking down a complex polynomial into simpler components, called factors. This is especially useful when solving polynomial equations because it allows for simplification.

For example, factorizing the polynomial \( x^n - y^n \) involves expressing it as a product of simpler polynomials such as \((x-y)(x^{n-1} + x^{n-2}y + ... + y^{n-1})\). Dividing by \((x-y)\) with no remainder, as shown by the Factor Theorem, confirms correct factorization.

When \( n \) is even or odd, additional factorization patterns, such as those involving \( x + y \), can apply. These factors can reveal symmetries within polynomials, simplify equations, and facilitate easier mathematical manipulation. Understanding these factorization techniques is key to analyzing polynomial behavior efficiently.

Polynomial Expressions

Polynomial expressions are mathematical phrases formed by variables raised to integer powers and combined through operations like addition, subtraction, and multiplication. They are foundational in algebra and appear in many mathematical contexts.

A basic polynomial expression can be represented in the general form \( a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \), where the exponents are whole numbers and the coefficients \( a_n, a_{n-1}, ..., a_0 \) are constants. The degree of the polynomial is the highest exponent present.

Understanding polynomial expressions and their properties is crucial when applying principles like the Factor Theorem or performing division and factorization. For instance, analyzing the expression \( x^n - y^n \) demands an understanding of the behaviors dictated by terms, powers, and symmetry. Mastery of polynomial expressions paves the way for solving complex algebraic problems effectively.