Problem 38 - Algebra 2

Question

a. Critical Thinking Simplify \(\frac{\left(2 x^{n}\right)^{2}-1}{2 x^{n}-1},\) where \(x\) is an integer and \(n\) is a positive integer. \((\text { Hint: Factor the numerator.) }\) b. Use the result from part (a) to show that the value of the given expression is always an odd integer.

Step-by-Step Solution

Answer

The simplified form of the given expression is \(2x^n + 1\), which is always an odd integer.

Step 1: Factor the numerator

Rewrite the expression as follows: \(\frac{(4x^{2n} - 1)}{2x^n - 1}\). Now, factor this as \(\frac{((2x^n - 1)(2x^n + 1))}{(2x^n - 1)}\) as per rule \(a^2 - b^2 = (a - b)(a + b)\), where \(a = 2x^n\) and \(b = 1\).

Step 2: Cancel out common factors

Notice that the term '2x^n - 1' occurs in both the numerator and the denominator. Therefore, we can cancel out the common factors in the numerator and the denominator which will simplify the expression to just \(2x^n + 1\).

Step 3: Show that the expression is always an odd integer

The expression has simplified to \(2x^n + 1\). If x is an integer and n is a natural number, \(x^n\) is an integer. 2 times an integer also gives an integer, on adding 1 to this product, the result is always an odd integer.

Key Concepts

Integer Properties

Integer properties are fundamental in understanding how numbers behave within operations like addition, multiplication, and factoring. Integers include positive and negative whole numbers, as well as zero. They do not include fractions or decimals.

When dealing with expressions, recognizing integer properties allows us to predict outcomes more easily. For example, any integer multiplied by 2 will always yield another integer, and when you add or subtract integers, the result will still be an integer.

• This property plays a crucial role in the given problem, as both \(x\) and \(x^n\) are assumed to be integers. Therefore, expressions like \(2x^n\) or \((2x^n + 1)\) remain as integers throughout the simplification process.
• Additionally, the problem asks us to show the expression simplifies to an odd integer. Knowing the integer property of even and odd, an even integer multiplied by 2 gives an even result, and adding 1 turns it into an odd integer.

Understanding these properties assists in the successful execution of operations and manipulations in algebraic expressions.

Factoring Quadratics

Factoring quadratics is a technique used to simplify expressions and solve equations. It involves rewriting a quadratic expression as a product of two binomials. In the given exercise, part of the process involves dealing with a quadratic form expressed within the numerator.

The expression \((4x^{2n} - 1)\) is a classic example where you can apply the difference of squares formula, \(a^2 - b^2 = (a - b)(a + b)\). Here, \(a\) is \(2x^n\) and \(b\) is 1. This factorization is key:

• By recognizing \(4x^{2n} - 1\) as a difference of squares, it factors into \((2x^n - 1)(2x^n + 1)\).
• Once factored, it becomes easier to simplify the expression by cancelling common terms with the denominator.

Factoring quadratics effectively reduces complex expressions into manageable pieces, enabling further simplification and understanding of the expression's properties.

Simplification Techniques

Simplification techniques in algebra involve reducing expressions or equations to their simplest form without changing their value. This makes them easier to interpret and solve. In the given problem, simplification plays a critical role in solving the expression.

Upon factoring the numerator, the expression \(\frac{(4x^{2n} - 1)}{2x^n - 1}\) transforms via factor cancellation as follows:

• With \((2x^n - 1)\) present in both the numerator and the denominator, these terms cancel each other out.
• This leaves us with \(2x^n + 1\), a much simpler form.

Cancelling common factors is a key simplification technique. It minimizes the expression's complexity without changing its original meaning or outcome.

The simplified expression \(2x^n + 1\) directly leads to proving the part of the problem about odd integers: doubling an integer and adding 1 guarantees an odd result, underscoring the effectiveness of simplification techniques.