Problem 51 - Elementary and Intermediate Algebra

Question

Factor. $$ 81 y^{2}-1 $$

Step-by-Step Solution

Answer

The factored form is \((9y - 1)(9y + 1)\).

Step 1: Identify the Expression as a Difference of Squares

The expression given is \(81y^2 - 1\). Notice that both terms are perfect squares. The first term \(81y^2 = (9y)^2\) and the second term \(1 = 1^2\). Because the expression is the difference of squares, \(a^2 - b^2\), it can be factored into \((a-b)(a+b)\).

Step 2: Recognize and Apply the Difference of Squares Formula

Using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\), identify \(a\) and \(b\) in the expression. Here, \(a = 9y\) and \(b = 1\). Substitute these into the formula: \(a^2 - b^2 = (a-b)(a+b)\).

Step 3: Factor the Expression

Substitute \(a = 9y\) and \(b = 1\) into the formula, giving the factors: \((9y - 1)(9y + 1)\). Thus, the original expression \(81y^2 - 1\) is factored into \((9y - 1)(9y + 1)\).

Key Concepts

Difference of Squares

The concept of the difference of squares is essential in algebra when it comes to factoring certain types of expressions. Simply put, when you have an expression that is structured as a square of one term minus the square of another term, it's called a difference of squares.

Let's break it down with a simple formula:

• The expression must be in the form of \(a^2 - b^2\)
• This represents the difference (or subtraction) of two perfect squares

The beauty of the difference of squares is that these expressions can be factored very easily by using the formula: \(a^2 - b^2 = (a-b)(a+b)\). This is particularly helpful because it breaks down complex expressions into simpler linear factors, making them much easier to work with.

For example, if you’re given: \(81y^2 - 1\), notice that both terms, \(81y^2\) and \(1\), are perfect squares. So, you can conclude that this expression is indeed a difference of squares. Factoring it becomes straightforward: recognize \(a = 9y\) and \(b = 1\), and apply the formula to get \((9y - 1)(9y + 1)\). By understanding how to identify and apply the difference of squares, you gain a powerful tool in factoring polynomials effectively.

Algebraic Expressions

Algebraic expressions form the foundation of algebra, consisting of variables, constants, and operators. Variables represent numbers whose values can change, like \(x\) or \(y\), while constants have a fixed value, such as 1, 9, or even more complex terms like \(81y^2\).

In algebra, expressions can be simplified, evaluated, and factored to reveal underlying relationships between numbers.

• Simplifying involves reducing expressions to their simplest form
• Evaluating means computing a value for the expression by substituting variables with numbers
• Factoring involves expressing a polynomial as a product of simpler polynomials

Consider the exercise \(81y^2 - 1\). It is an algebraic expression composed of two main terms: \(81y^2\) and \(1\). Each term can potentially have variables, as seen with \(y\), or just be constants. By utilizing factoring techniques, specifically the difference of squares in this case, the algebraic expression is broken down into two simpler binomials \((9y - 1)\) and \((9y + 1)\). Simplifying algebraic expressions this way is not only about finding simpler forms but also about making complex problems more manageable and solvable.

Factoring Techniques

Factoring is a crucial skill in algebra that involves breaking down expressions into products of other simpler expressions. Mastering factoring techniques is like having a toolbox filled with strategies to tackle different types of polynomials.

There are various techniques for factoring, including:

• Common Factor: Identifying and taking out the greatest common factor of all terms in the expression
• Difference of Squares: Recognizing expressions formed by the subtraction of two squares and applying the formula
• Trinomials: Factoring quadratic expressions like \(ax^2 + bx + c\)

When given the expression \(81y^2 - 1\), which is a perfect case for the difference of squares, the process becomes even smoother. Recognizing the structure as \(a^2 - b^2\) allows you to quickly apply the difference of squares technique to write the expression as \((9y - 1)(9y + 1)\).

Practicing different factoring techniques equips you with the ability to simplify and solve polynomial expressions effectively, setting the stage for more advanced algebraic problem-solving. Being proficient in these methods not only aids in cracking algebraic expressions but also contributes to a solid foundation in algebra.