Problem 103 - Algebra and Trigonometry

Question

Factoring An 1 Verify the factoring formulas in the list by expanding and simplifying the right-hand side in each case. $$ \begin{array}{l}{A^{2}-1=(A-1)(A+1)} \\ {A^{3}-1=(A-1)\left(A^{2}+A+1\right)} \\\ {A^{4}-1=(A-1)\left(A^{3}+A^{2}+A+1\right)}\end{array} $$ On the basis of the pattern displayed in this list, how do you think \(A^{5}-1\) would factor? Verify your conjecture. Now generalize the pattern you have observed to obtain a factorization formula for \(A^{n}-1,\) where \(n\) is a positive integer.

Step-by-Step Solution

Answer

The pattern is \(A^n - 1 = (A-1)(A^{n-1} + \\dots + 1)\).

Step 1: Understand the Problem

We are asked to verify the factorization formula for different powers of \(A\) and observe any patterns. Let's start by expanding the given factorization formulas and see if they match with the left side of the equations.

Step 2: Verify \(A^2 - 1\)

Expand \((A-1)(A+1)\):\[(A - 1)(A + 1) = A^2 + A - A - 1 = A^2 - 1\]This matches the left side, verifying the formula is correct.

Step 3: Verify \(A^3 - 1\)

Expand \((A-1)(A^2 + A + 1)\):\[(A-1)(A^2 + A + 1) = A^3 + A^2 + A - A^2 - A - 1 = A^3 - 1\]This matches the left side, verifying the formula is correct.

Step 4: Verify \(A^4 - 1\)

Expand \((A-1)(A^3 + A^2 + A + 1)\):\[(A-1)(A^3 + A^2 + A + 1) = A^4 + A^3 + A^2 + A - A^3 - A^2 - A - 1 = A^4 - 1\]This matches the left side, verifying the formula is correct.

Step 5: Conjecture for \(A^5 - 1\)

Based on the pattern,\[A^5 - 1 = (A-1)(A^4 + A^3 + A^2 + A + 1)\].

Step 6: Verify Conjecture for \(A^5 - 1\)

Expand \((A-1)(A^4 + A^3 + A^2 + A + 1)\):\[(A-1)(A^4 + A^3 + A^2 + A + 1) = A^5 + A^4 + A^3 + A^2 + A - A^4 - A^3 - A^2 - A - 1 = A^5 - 1\]This matches the left side, verifying our conjecture is correct.

Step 7: General Pattern for \(A^n - 1\)

Observe that \[A^n - 1 = (A-1)(A^{n-1} + A^{n-2} + \dots + 1)\]In general, the factorized form of \(A^n - 1\) is \[(A-1)(A^{n-1} + A^{n-2} + \dots + 1)\].

Key Concepts

Verification of Factoring Formulas

When it comes to factoring formulas, verification is key. To verify a factoring formula, we need to expand the product on the right-hand side and check if it matches the original expression on the left-hand side. This ensures that our factorization is accurate.
For example, consider the simpler expressions:

• For \(A^2 - 1\), when we expand \((A-1)(A+1)\), it results in \(A^2 + A - A - 1 = A^2 - 1\). This confirms that the formula \(A^2 - 1 = (A-1)(A+1)\) is valid.


• Similarly, for \(A^3 - 1\), expanding \((A-1)(A^2 + A+1)\) results in \(A^3 + A^2 + A - A^2 - A - 1 = A^3 - 1\), verifying this formula too.
  

Verification helps us trust the formulas we're using and deepens our understanding of the algebraic operations at play.

Pattern Recognition in Algebra

Recognizing patterns is an incredibly useful skill in algebra, especially when it comes to factoring. In this exercise, we observe a repetitive structure as we factor higher powers of \(A\).
For the expressions \(A^2 - 1\), \(A^3 - 1\), and \(A^4 - 1\), the factors feature the term \(A - 1\), followed by a decrease in powers of \(A\) in the other factor.

• For each power, the additional factor consists of terms where the powers decrease by one each step: \(A^{n-1} + A^{n-2} + \ldots + 1\).
  
• These patterns not only help us form conjectures for larger powers, such as \(A^5 - 1\), but also assist in generalizing the expression for any power \(n\).
  

Recognizing these patterns is like unraveling a mystery; it guides us to concise solutions.

Powers and Exponents

Understanding powers and exponents is crucial when dealing with polynomial factorization. Powers denote repeated multiplication of a number, and exponents tell us how many times to multiply that number by itself.

• In our examples, the main expression \(A^n - 1\) represents a polynomial where \(A\) is raised to the power of \(n\), minus 1.
  
• Expanding using the distributive property shows how different exponents combine and cancel out terms, simplifying the expression back to its original power of \(n\).
  
• Handling exponents requires us to distribute powers across terms, maintaining balance to ensure terms either contribute to or cancel out in the expansion.
  

Dealing with powers help us manipulate algebraic expressions effectively, aiding both in solving equations and verifying identities.

Generalizing Mathematical Patterns

By generalizing mathematical patterns, we translate specific examples into universal formulas. In this context, identifying a consistent method across examples allows us to create a formula valid for any positive integer \(n\).

• Our exercise started with specific cases, such as \(A^3 - 1\) and progressed to \(A^5 - 1\).
  
• Observing the pattern led us to formulate \(A^n - 1 = (A-1)(A^{n-1} + A^{n-2} + \ldots + 1)\), where each term reduces in power by one until reaching the constant
  1.
• This generalization simplifies problem-solving, as we apply a tested approach to any similar expression.
  

Generalizing allows us to tackle complex problems efficiently, offering a strategic solution without retracing steps for each new power.