Problem 28 - Algebra and Trigonometry

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

The polynomial \(x^{2n-1} + y^{2n-1}\) is divisible by \(x+y\) for any natural number \(n\).

Step 1: Understand the Problem

We want to demonstrate that \(x + y\) is a factor of \(x^{2n-1} + y^{2n-1}\) for any natural number \(n\). This means that \(x^{2n-1} + y^{2n-1}\) can be expressed as \((x+y)Q(x,y)\) for some polynomial \(Q(x,y)\).

Step 2: Use Polynomial Long Division

To show that \(x + y\) divides \(x^{2n-1} + y^{2n-1}\), let's consider dividing \(x^{2n-1} + y^{2n-1}\) by \(x + y\) using polynomial long division. Polynomial long division will help find the remainder when \(x^{2n-1} + y^{2n-1}\) is divided by \(x + y\).

Step 3: Analyze Remainder Theorem

According to the Remainder Theorem, if \(f(x)\) is a polynomial and \(f(x)\) is divided by \(x+a\), the remainder is \(f(-a)\). Here, \(f(x) = x^{2n-1} + y^{2n-1}\) and we set \(x = -y\). Thus, we need to evaluate \((-y)^{2n-1} + y^{2n-1}\), which simplifies to \(-y^{2n-1} + y^{2n-1} = 0\).

Step 4: Conclude No Remainder

Since evaluating \(f(-y)\) leads to 0, there is no remainder, meaning \(x+y\) divides \(f(x) = x^{2n-1} + y^{2n-1}\) without remainder. Thus, \(x + y\) is indeed a factor of \(x^{2n-1} + y^{2n-1}\).

Key Concepts

Remainder Theorem

The Remainder Theorem connects polynomial division with evaluations of polynomials. It states that if a polynomial \(f(x)\) is divided by \(x+a\), the remainder of this division is simply \(f(-a)\). This is a powerful tool as it allows you to quickly determine the remainder by just substituting a value.

To see it in action, let's apply it to the given exercise. We have \(f(x) = x^{2n-1} + y^{2n-1}\) and need to determine if \(x + y\) is a factor. Using the Remainder Theorem, we compute \(f(-y)\). This means replacing every occurrence of \(x\) in the polynomial with \(-y\), resulting in:

\((-y)^{2n-1} + y^{2n-1}\).

This simplifies to \(-y^{2n-1} + y^{2n-1}\). Since this equals zero, there is no remainder, confirming that \(x + y\) is indeed a factor.

Polynomial Long Division

Polynomial long division is an essential method, similar to numerical long division, used to divide polynomials. This technique is particularly helpful for dividing more complex polynomials, finding quotients, and determining remainders, if any.

To divide \(x^{2n-1} + y^{2n-1}\) by \(x+y\) using polynomial long division, follow these steps:

• Set up the division by writing \(x+y\) outside the division bracket, and \(x^{2n-1} + y^{2n-1}\) inside the bracket.
• Determine how many times the leading term of \(x+y\), which is \(x\), fits into the leading term of the dividend \(x^{2n-1}\).
• Multiply \(x+y\) by this term and subtract it from the dividend.
• Repeat the process with the new polynomial that results after subtraction, until no terms are left or the degree of the remaining polynomial is smaller than \(x+y\).

When complete, the process will show if there is a remainder or not. If the remainder is zero, as it is in our example, \(x+y\) is a factor of the original polynomial.

Factor Theorem

The Factor Theorem is closely related to the Remainder Theorem and is a fundamental concept in polynomial algebra. It provides a straightforward method to determine when a polynomial is divisible by a linear factor \(x - a\). Specifically, it states that \(x - a\) is a factor of \(f(x)\) if and only if \(f(a) = 0\).

Applying this to our situation, we set \(x=-y\) in the polynomial \(x^{2n-1} + y^{2n-1}\). If \(x+y\) is a factor, substituting \(-y\) should yield zero. As calculated earlier, substituting gives us \(-y^{2n-1} + y^{2n-1} = 0\).

This zero result affirms that \(x + y\) divides the polynomial perfectly, illustrating the Factor Theorem in action. It’s a quick and efficient way to confirm factors without having to fully perform polynomial division each time.

Remember, this powerful tool assists in identifying factors and simplifying polynomial expressions, making it especially useful for verifying factorization and roots in algebra.