Problem 64 - Algebra 2

Question

Factor each expression completely. $$ \frac{1}{2} x^{2}-\frac{1}{2} $$

Step-by-Step Solution

Answer

\( \frac{1}{2}(x + 1)(x - 1) \)

Step 1: Factor out the Greatest Common Factor (GCF)

Identify and factor out the greatest common factor from the terms in the expression. In this case, the GCF is \( \frac{1}{2} \).

Step 2: Recognize a Difference of Squares

After factoring out the GCF, recognize that the remaining terms form a difference of squares pattern \( a^2 - b^2 \) which can be factored into \( (a + b)(a - b) \) where \( a = x \) and \( b = 1 \) in this context.

Step 3: Apply the Difference of Squares Formula

Factor the difference of squares using the formula \( (a + b)(a - b)\) to obtain the completely factored expression.

Key Concepts

Factoring Polynomials

Factoring polynomials is a key skill in algebra that involves breaking down a polynomial into a product of simpler polynomials. Think of it like finding the original pieces of a Lego structure. It's the process of transforming a complicated expression into a multiplication of simpler ones, which often reveal interesting properties and can simplify calculations. When factoring, we look for patterns such as the difference of squares, common factors, or special binomial products. The goal is to rewrite the polynomial as a product where each factor is simpler than the original expression.

For instance, with the expression \( \frac{1}{2} x^{2}-\frac{1}{2} \), we start by searching for a common factor in each term. Once the common factor is out of the way, the structure of the expression becomes clearer, and if patterns like the difference of squares emerge, we use corresponding formulas to factor further.

Greatest Common Factor (GCF)

The Greatest Common Factor (GCF) is the largest factor shared by two or more numbers or terms. In algebra, it is useful for factoring expressions, simplifying fractions, and solving equations. The key to factoring any algebraic expression often starts with identifying the GCF. It's like uncovering the biggest building block common to all parts of a structure.

For the expression given, \( \frac{1}{2} x^{2}-\frac{1}{2} \), we spot that both terms have \( \frac{1}{2} \) in common. Factoring it out is our first step, similar to decluttering a space to see the layout more clearly. Once the GCF is factored out, we can more easily identify other factoring techniques to apply, such as recognising the difference of squares.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations that represent a specific mathematical idea. Just like phrases in a language, they convey information without an equals sign — unlike equations. These expressions can have multiple terms and powers and can be as simple or as complex as the combination of its parts.

In our exercise, \( \frac{1}{2} x^{2}-\frac{1}{2} \) is an algebraic expression with two terms. The terms are connected by subtraction, making it a binomial (bi- meaning two). The expression includes constants (\( \frac{1}{2} \)), a variable (\(x\)), and an exponent (the square in \(x^{2}\)), demonstrating the variety of components that can be involved. The goal in algebra often involves manipulating these expressions in useful ways, such as factoring, to explore and solve different mathematical scenarios.

Difference of Squares Formula

The difference of squares formula is a specific factoring pattern that applies when you have two perfect squares separated by a subtraction sign. The pattern is written as \( a^2 - b^2 = (a + b)(a - b) \). It reveals that any expression structured as one square minus another square can be broken into the product of two binomials — one with a sum and one with a difference.

After factoring out the GCF in our problem, we're left with \( x^{2}-1 \), which fits this pattern perfectly, since \(x^{2}\) and \(1\) are both perfect squares. Using the difference of squares formula, we can factor the expression into \( (x + 1)(x - 1) \). This simple yet powerful formula allows us to transform expressions and can be particularly helpful in solving equations and simplifying algebraic expressions.