Problem 26 - Precalculus: Mathematical for Calculus

Question

Show that \(x+y\) is a factor of \(x^{2 n-1}+y^{2 n-1}\) for all natural numbers \(n\)

Step-by-Step Solution

Answer

Yes, \(x+y\) is a factor of \(x^{2n-1} + y^{2n-1}\) for all natural numbers \(n\).

Step 1: Understand the Problem

We need to show that \(x+y\) is a factor of the polynomial \(x^{2n-1} + y^{2n-1}\) for any natural number \(n\). This means we must demonstrate that when \(x^{2n-1} + y^{2n-1}\) is divided by \(x+y\), the remainder is zero.

Step 2: Use Polynomial Division

Consider the polynomial \(P(x, y) = x^{2n-1} + y^{2n-1}\). To show \(x+y\) is a factor, we perform polynomial long division of \(P(x, y)\) by \(x+y\). By the Factor Theorem, \(x+y\) is a factor if \(P(-y, y) = 0\).

Step 3: Apply the Factor Theorem

Substitute \(x = -y\) into the polynomial \(P(x, y) = x^{2n-1} + y^{2n-1}\). This gives \((-y)^{2n-1} + y^{2n-1}\).

Step 4: Simplify the Expression

Simplify \((-y)^{2n-1} + y^{2n-1}\) by factoring out \(y^{2n-1}\): \[(-1)^{2n-1} y^{2n-1} + y^{2n-1} = -y^{2n-1} + y^{2n-1} = 0.\]

Step 5: Conclude That x+y is a Factor

Since substituting \(x = -y\) yields a result of zero, by the Factor Theorem, \(x+y\) is indeed a factor of \(x^{2n-1} + y^{2n-1}\) for all natural numbers \(n\).

Key Concepts

Polynomial Division

Polynomial division is a method to simplify polynomials by dividing one polynomial by another. The goal is to find the quotient and remainder when the division is carried out. This process is akin to long division with numbers, just a bit more complex due to the presence of variables.Imagine you have a polynomial, like the one in our exercise, which we denote as \( P(x, y) = x^{2n-1} + y^{2n-1} \). We want to check whether \( x+y \) is a factor, which means that dividing \( P(x, y) \) by \( x+y \) should give a remainder of zero.To perform a polynomial division, follow these steps:- **Arrange the polynomials**: Make sure both polynomials are in the standard form, with terms ordered in descending powers.- **Divide the first term**: Divide the highest degree term of the dividend by the highest degree term of the divisor.- **Multiply and subtract**: Multiply the entire divisor by the result of the previous step and subtract from the dividend.- **Repeat**: Continue this process with the new polynomial that results.Polynomial division is an effective tool for simplifying expressions and plays a crucial role in revealing whether a given polynomial is a factor of another.

Factor Theorem

The Factor Theorem is a fundamental concept in algebra, particularly useful when working with polynomials. It states that a polynomial \( P(x) \) has a factor \((x-a)\) if and only if \( P(a) = 0 \), meaning the value of the polynomial at \( x = a \) is zero.In our example, the Factor Theorem is applied as follows:- We need to test if \( x+y \) is a factor of \( x^{2n-1} + y^{2n-1} \).- We substitute \( x = -y \) into the polynomial giving us \( (-y)^{2n-1} + y^{2n-1} \).- Simplifying, we observe that this equals zero, as both terms cancel each other out.Therefore, according to the Factor Theorem, \( x+y \) is indeed a factor since the polynomial evaluated at \( x = -y \) results in zero. This is incredibly helpful for quickly verifying factors without performing long division.

Natural Numbers

Natural numbers are the set of positive integers beginning from 1 and continuing infinitely: 1, 2, 3, and so on. They are the basic building blocks in number theory and contribute greatly to understanding mathematical concepts.In the context of polynomial factorization, natural numbers, denoted usually by the letter \( n \), often represent exponents or coefficients. In our given polynomial \( x^{2n-1} + y^{2n-1} \), the exponent \( 2n-1 \) tells us how high the degree of each term is.Here's how natural numbers influence the polynomial:- **Exponentiation**: The term \( 2n-1 \) reveals the order of each term in the polynomial. As \( n \) increases, the exponent grows, leading to more complex terms.- **Finiteness**: Using \( n \) ensures terms are finite and manageable within mathematical resources, helping maintain structure and balance in expressions.Understanding the role of natural numbers in polynomials is vital as it affects both the complexity of calculation and the factorization process itself.