Problem 49 - Elementary and Intermediate Algebra

Question

Factor by first grouping the appropriate terms. \(x^{2}+12 x+36-y^{2}\)

Step-by-Step Solution

Answer

The expression factors to \((x + 6 - y)(x + 6 + y)\).

Step 1: Recognize a Pattern

Notice that the expression \(x^2 + 12x + 36\) resembles a perfect square trinomial. This can be factored as \((x + a)^2\), where \(a^2 = 36\) and \(2a = 12\). Thus, \(a = 6\). Write the expression as \((x + 6)^2 - y^2\).

Step 2: Apply the Difference of Squares Formula

Recognize that \((x + 6)^2 - y^2\) is a difference of squares, which can be factored using the formula \(a^2 - b^2 = (a - b)(a + b)\). Here, \(a = x + 6\) and \(b = y\).

Step 3: Factor the Expression

Apply the difference of squares formula: \((x + 6) - y\) and \((x + 6) + y\). Hence, the expression factors to \((x + 6 - y)(x + 6 + y)\).

Key Concepts

Perfect Square Trinomial

Factoring a perfect square trinomial is like finding a beautiful pattern hidden within the numbers. Trinomials, which are algebraic expressions with three terms, sometimes present themselves as perfect squares. What does this mean? Essentially, a perfect square trinomial can be neatly expressed as a square of a binomial. For example, let's consider the trinomial \(x^2 + 12x + 36\). With a little observation, you can spot its special arrangement and factor it accordingly.

• Look for a trinomial that can be expressed in the form \((x + a)^2\).
• The first term, \(x^2\), is identified as a perfect square, \(x\) being squared.
• The constant term \(36\) is also a perfect square, since \(6^2 = 36\).
• To confirm, check the middle term: it should match \(2a = 12\) when \(a = 6\).

This tells you that the trinomial is indeed a perfect square and can be effortlessly rewritten as \((x + 6)^2\). Each part of your trinomial ties up into a simple, elegant solution: you’ve packed it neatly into a square!

Difference of Squares

The difference of squares is a remarkable algebraic identity that often reveals itself once you learn to look for it. When you see two perfect squares separated by a subtraction sign, you’re in the realm of the difference of squares formula, which is written as \(a^2 - b^2 = (a - b)(a + b)\). This formula is incredibly helpful in algebra for simplifying and solving problems.

• Identify your expressions \(a^2\) and \(b^2\).
• In our problem, \((x + 6)^2 - y^2\), \(a\) is \(x + 6\), and \(b\) is \(y\).
• The difference of squares tells us to form two binomials: one by subtracting \(b\) from \(a\), and the other by adding \(b\) to \(a\).

Applying the formula, the expression \((x + 6)^2 - y^2\) transforms into \((x + 6 - y)(x + 6 + y)\). It’s a clear, powerful way to make expressions less complex, and it's all about spotting the simplicity amid the structure.

Algebraic Expressions

Algebraic expressions are the backbone of algebra, where letters and numbers work together like notes in a symphony. Designed to represent relationships and solve problems, they often involve variables (like \(x\) and \(y\)), constants, and operations (like addition and subtraction). As you become familiar with these expressions, you'll start to see their inherent patterns.

• Each term in an expression contributes to its overall form and function.
• Expressions can often be rewritten and simplified, leading to easier solutions.
• In our work with \(x^2 + 12x + 36 - y^2\), understanding how each term interacts is key.

The power of algebraic expressions lies in their flexibility. By recognizing their components and relationships, you can manipulate them effectively. When you practice and understand them, you become adept at revealing simpler parts within seemingly complex expressions, making problem-solving a much more expressive and satisfying experience.