Problem 94 - Algebra A Combined Function

Question

\(\left(x^{2}+2 x+1\right)-36 y^{2}\) (Hint: Factor the trinomial in parentheses first.)

Step-by-Step Solution

Answer

The factored form is \((x+1-6y)(x+1+6y)\).

Step 1: Recognize the Structure

Notice that the expression inside the parentheses, \(x^2 + 2x + 1\), is a perfect square trinomial. It can be rewritten as \((x+1)^2\).

Step 2: Substitute and Rewrite

Substitute the perfect square trinomial with its factored form. The expression becomes \((x+1)^2 - 36y^2\).

Step 3: Identify Further Factorization

Recognize that the expression is now structured as a difference of squares: \((a^2 - b^2)\), where \(a = (x+1)\) and \(b = 6y\).

Step 4: Apply the Difference of Squares Formula

Use the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\). Substitute \(a\) and \(b\) with \((x+1)\) and \(6y\), respectively. Thus, the expression becomes \((x+1-6y)(x+1+6y)\).

Key Concepts

Understanding Perfect Square Trinomials

A perfect square trinomial is a type of algebraic expression that arises from squaring a binomial. This means if you expand the binomial

• ext{(a + b)}^2 ightarrow a^2 + 2ab + b^2

Using this identity, you recognize perfect square trinomials by their structure: the first term and the last term are perfect squares, and the middle term is twice the product of the square roots of those terms.

For example, in our expression \(x^2 + 2x + 1\), the first term \(x^2\) is \(x^2\), the last term \(1\) is \(1^2\), and the middle term \(2x\) is \(2 imes x imes 1\). Therefore, we factor it as \( (x+1)^2 \).

Understanding how to spot and factor perfect square trinomials is valuable because it simplifies the expression and prepares it for further operations.

Recognizing Difference of Squares

The difference of squares is a pattern associated with binomials where \(a^2 - b^2\) can be expressed as \((a-b)(a+b)\). This pattern involves two squared terms with a subtraction sign between them. It's called "difference of squares" because that's exactly what it is: a difference (subtraction) of two squares.

In the expression \( (x+1)^2 - 36y^2 \), we can see that it is indeed a difference of squares. Here, \(a = x+1\) and \(b = 6y\), since \( (x+1)^2 \) and \( (6y)^2 \) are perfect squares.

Recognizing this structure allows us to apply the formula for factoring them, \( (x+1)^2 - (6y)^2 = (x+1-6y)(x+1+6y) \). This method is powerful, as it quickly transforms complex expressions into simpler products of factors.

Effective Factoring Techniques

Factoring is a cornerstone of algebra which involves rewriting an expression as a product of its factors. It can simplify equations, making them easier to solve, especially when dealing with polynomials.

In our example, the initial expression is \( \left(x^{2}+2x+1\right)-36y^{2} \). With the application of factoring techniques, we used two main patterns:

• Perfect Square Trinomial: \( x^2 + 2x + 1 = (x+1)^2 \)
• Difference of Squares: \((x+1)^2 - (6y)^2 = (x+1-6y)(x+1+6y)\)

Each step in factoring breaks down the expression into simpler parts, making it not only easier to handle but also easier to understand and solve further problems involving these expressions.

Mastery of factoring techniques paves the way for solving more complex algebraic equations, making it a vital skill in any math curriculum.