Problem 50 - Algebra and Trigonometry

Question

31–76 ? Factor the expression completely. $$ 4 x^{2}-25 $$

Step-by-Step Solution

Answer

The expression factors to \((2x + 5)(2x - 5)\).

Step 1: Identify the Structure

First, recognize that the expression \( 4x^2 - 25 \) resembles a difference of squares. A difference of squares is expressed as \( a^2 - b^2 \), which factors into \((a + b)(a - b)\).

Step 2: Rewrite as a Difference of Squares

Notice that \( 4x^2 \) can be rewritten as \((2x)^2\) and \( 25 \) as \(5^2\). Thus, the expression becomes \((2x)^2 - 5^2\).

Step 3: Apply Difference of Squares Formula

Using the difference of squares formula \( a^2 - b^2 = (a + b)(a - b) \), we can factor \( (2x)^2 - 5^2 \) as \((2x + 5)(2x - 5)\).

Step 4: Verify the Factorization

To ensure that \((2x + 5)(2x - 5)\) is the correct factorization, multiply the factors back together. \[(2x + 5)(2x - 5) = 4x^2 - 10x + 10x - 25\] Simplify to \(4x^2 - 25\), confirming the factorization is correct.

Key Concepts

Difference of Squares

The concept of a "difference of squares" is a fundamental pattern in algebra that can simplify the process of factoring certain types of expressions. When expressions are of the form \(a^2 - b^2\), they can be factored using a straightforward formula:

• \(a^2 - b^2 = (a + b)(a - b)\)

This formula capitalizes on the symmetrical properties of squares and exploits the fact that the middle terms cancel when expanded. For example, to see how this applies, consider \(4x^2 - 25\). First, recognize that both terms are perfect squares, with

• \(4x^2 = (2x)^2\)
• \(25 = 5^2\)

This reveals the expression as a difference of squares. Using our formula, \(4x^2 - 25\) becomes \((2x + 5)(2x - 5)\). Always checking by expanding the factors back is a good practice to ensure accuracy.

Polynomial Factorization

Polynomial factorization involves breaking down a polynomial into simpler, non-divisible components, often called factors. This process is essential in simplifying expressions, solving equations, and analyzing polynomial functions. The key with polynomial factorization is identifying patterns or techniques, such as the difference of squares, to break down complex expressions into manageable parts.
The process can include various approaches, such as

• Factoring by grouping
• Factoring trinomials with trial and error
• Recognizing special products

In the example \(4x^2 - 25\), recognizing the difference of squares allowed for the use of the special products method, directly applying the formula \((a + b)(a - b)\). For other polynomials, finding common factors or employing the quadratic formula is often necessary. Factoring not only simplifies expressions but also aids in graphing polynomials and solving algebraic equations.

Algebraic Expressions

Algebraic expressions are combinations of variables, constants, and operators (such as addition and multiplication). Understanding how to manipulate these expressions is a cornerstone of algebra. Within an algebraic expression, each part has a name:

• Terms: Parts of the expression separated by addition or subtraction.
• Coefficients: Numbers multiplying the variables.
• Constants: Fixed numbers that don't change.

When tackling exercises involving algebraic expressions, it's crucial to understand the objective, whether it's simplifying, factoring, or solving for a variable. The problem \(4x^2 - 25\) illustrates how algebraic expressions can be manipulated through recognizing patterns, such as the difference of squares.
Being adept at transforming these expressions allows you to know which techniques, like distributing, combining like terms, or factoring, will bring about the desired result. Mastering algebraic expressions is foundational for more advanced mathematics, as it forms the basis for function analysis, calculus, and beyond.