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Chapter 6: Problem 34

Factor the given expressions completely. $$(a-b)^{2}-1$$

Short Answer

Expert verified
The expression factors to \((a-b-1)(a-b+1)\).

Step by step solution

01

Recognizing a Difference of Squares

The given expression is \((a-b)^{2}-1\). Notice that this expression is in the form of a difference of squares, where \((a-b)^2\) is a perfect square and \(1\) is also a perfect square, as it equals \(1^2\). We can apply the difference of squares formula: \(x^2 - y^2 = (x-y)(x+y)\).
02

Apply the Difference of Squares Formula

Using the formula for the difference of squares, we identify \( (a-b)^2 \) as \(x^2\) and \(1\) as \(y^2\). This allows us to write:\[ (a-b)^{2} - 1 = ig((a-b) - 1ig)ig((a-b) + 1ig) \]Thus, the expression factors to \( ((a-b) - 1)((a-b) + 1) \).
03

Simplifying the Factors

Simplify each term inside the factored expression. The expression becomes:\[ (a-b-1)(a-b+1) \]These cannot be simplified further and represent the factors of the original expression.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Squares
The difference of squares is a unique algebraic identity where two perfect squares are being subtracted. It follows the formula \(x^2 - y^2 = (x-y)(x+y)\). This formula is significant because it allows us to factor expressions that at first glance may seem complex. The "difference" means we're subtracting, and the "squares" refer to numbers or expressions multiplied by themselves. Recognizing this pattern is crucial because it transforms the expression into a more manageable pair of binomials. In our initial exercise,
  • the expression \( (a-b)^{2}-1 \) can be viewed as two perfect squares.
  • The square of \( (a-b) \) and the square of 1, which is \( 1^2 \).
  • This recognition guides us in applying the difference of squares formula effectively.
Spotting a difference of squares can simplify the process of factoring, making it easier to resolve the expression into its component parts.
Factoring Expressions
Factoring expressions involves breaking down a complex expression into simpler components, or 'factors', that when multiplied together give back the original expression. One of the crucial strategies in factoring is identifying patterns such as common factors, trinomials, or in this case, the difference of squares. When approaching an expression like
  • \((a-b)^2 - 1\), we seek to express it in a form that reveals its factors.
  • Using the difference of squares pattern, we write it as \( ((a-b) - 1)((a-b) + 1) \).
  • This breaks the expression into two linear binomials.
Factoring makes it easier to solve equations, manage polynomials, and provides a foundation for more advanced algebra topics.
Perfect Squares
A perfect square is the result of a number or an algebraic expression being multiplied by itself. They are essential in algebra because they frequently appear in expressions that need to be simplified or factored.In the context of the given exercise,
  • \((a-b)^2\) is a perfect square as it denotes \( (a-b) \) multiplied by itself.
  • Furthermore, the number 1 is also considered a perfect square, represented as \( 1^2 \).
When factoring expressions, identifying perfect squares provides clues about the possible methods of factoring. Recognizing these squares led us to use the difference of squares method, effectively decomposing the initial expression into simpler, manageable factors.

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Most popular questions from this chapter

Factor the given expressions completely. Each is from the technical area indicated. $$3 A d u^{2}-4 A d u v+A d v^{2} \quad \text { (water power) }$$

In Exercises \(29-64,\) reduce each fraction to simplest form. $$\frac{(2 x-1)(x+6)}{(x-3)(1-2 x)}$$

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