Problem 28 - Precalculus Mathematics 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

\(x+y\) is a factor of \(x^{2n-1}+y^{2n-1}\).

Step 1: Identify the Problem

We need to prove that \(x+y\) divides \(x^{2n-1} + y^{2n-1}\). For this, we have to show that if \(x = -y\), then \(x^{2n-1} + y^{2n-1} = 0\). This will demonstrate \(x+y\) is a factor.

Step 2: Substitute x with -y

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

Step 3: Simplify the Expression

Simplify \((-y)^{2n-1} + y^{2n-1}\). Note that since \(2n-1\) is odd, \((-y)^{2n-1} = -(y^{2n-1})\). Therefore, the expression becomes \(-(y^{2n-1}) + y^{2n-1}\).

Step 4: Calculate the Result

Calculate \(-(y^{2n-1}) + y^{2n-1}\). This results in \(0\). Hence, when \(x = -y\), \(x^{2n-1} + y^{2n-1} = 0\).

Step 5: Conclude the Solution

Since \(x^{2n-1} + y^{2n-1}\) equals zero when \(x = -y\), it shows that \(x+y\) is a factor of the original expression for all natural numbers \(n\).

Key Concepts

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operators. At their core, these are formed by combining constants, like 3 or 5, and variables, such as \(x\) or \(y\), using operations like addition, subtraction, multiplication, and division.
For example, in the exercise where the expression is \(x^{2n-1} + y^{2n-1}\), \(x\) and \(y\) are variables, and they can take on different values. Meanwhile, \(2n-1\) represents an exponent which depends on the natural number \(n\).

• A key point in manipulating algebraic expressions is understanding properties of exponents, like how powers of \(y\) behave.
• Another crucial aspect is recognizing patterns, such as symmetry, which can simplify complex problems.

By understanding how these expressions are structured and how their components interact, one can develop strategies to simplify and factor them, as shown in the solution.

Remainder Theorem

The Remainder Theorem is a useful tool when dealing with polynomial division. This theorem states that for a polynomial \(f(x)\), when divided by a binomial of the form \(x - c\), the remainder of the division is simply \(f(c)\).
In essence, this method allows us to evaluate the remainder without performing lengthy division.

• For example, if \(f(x) = x^2 + 3x + 2\) and you want to check if \(x - 1\) is a factor, calculate \(f(1)\).
• If \(f(1) = 0\), then \(x - 1\) is indeed a factor.

In our given exercise, even though we didn’t formally apply the remainder theorem, the process of substitution and simplification effectively showed that \(x+y\) was a factor, akin to making the remainder zero by setting \(x = -y\).

Substitution Method

The Substitution Method in algebra involves replacing a variable with a given value or expression to simplify the problem. This can unravel intricate algebraic expressions, making them easier to manipulate and solve.
In the original solution, we used the substitution \(x = -y\) to explore whether \(x+y\) is a factor of the expression \(x^{2n-1} + y^{2n-1}\).

• By substituting \(x = -y\), the expression transforms into \((-y)^{2n-1} + y^{2n-1}\).
• This change exploits the property of exponents and helps highlight patterns, like terms cancelling out.

The simplification that follows stemmed directly from the substitution made, leading seamlessly to the conclusion that \(x+y\) divides the expression for all natural numbers \(n\) by achieving a result of zero.