Problem 50 - Elementary and Intermediate Algebra

Question

Factor by first grouping the appropriate terms. \(x^{2}-6 x+9-4 y^{2}\)

Step-by-Step Solution

Answer

The factored form is \((x-3-2y)(x-3+2y)\).

Step 1: Recognize the Expression

The expression to be factored is given as \(x^{2}-6x+9-4y^{2}\). Start by recognizing that part of this expression is a trinomial, \(x^{2}-6x+9\), and the other part, \(-4y^{2}\), is a difference of squares.

Step 2: Factor the Trinomial

Identify that \(x^{2} - 6x + 9\) is a perfect square trinomial. It can be factored into \((x-3)^2\) because \( (x-3)(x-3) = x^{2} - 6x + 9 \).

Step 3: Recognize and Factor the Difference of Squares

Notice that \( (x-3)^2 - 4y^2 \) is a difference of squares. The difference of squares formula is \(a^2 - b^2 = (a-b)(a+b)\). Here, \(a = x-3\) and \(b = 2y\).

Step 4: Apply Difference of Squares Formula

Apply the difference of squares formula: \((x-3)^2 - (2y)^2 = ((x-3) - 2y)((x-3) + 2y)\).

Step 5: Simplify Final Expression

The expression can be rewritten as \((x-3-2y)(x-3+2y)\). Thus, the factored form of \(x^2 - 6x + 9 - 4y^2\) is final.

Key Concepts

Understanding Trinomials

A trinomial is a type of polynomial that consists of three distinct terms. In many cases, trinomial expressions can be factored into simpler binomial parts. This process is similar to looking for pairs of numbers or terms that multiply and add up to form the coefficients and terms of the original trinomial.
Let's take the given expression from the exercise:

• The trinomial here is \(x^2 - 6x + 9\).

To factor this trinomial, we need to identify patterns such as perfect square trinomial forms or seek combinations of terms that simplify the expression.
The term \(x^2 - 6x + 9\) fits the pattern of a perfect square trinomial, which leads us to factor it into \((x-3)(x-3)\) or simply \((x-3)^2\). The steps involve recognizing that the middle term, \(-6x\), can be expressed as twice the product of \(x\) and \(-3\). As a result, this trinomial is a great example where recognizing the perfect square pattern makes factoring much easier.

Exploring the Difference of Squares

The concept of the "difference of squares" is a special algebraic identity that simplifies expressions of the form \(a^2 - b^2\).
This identity states:

• \(a^2 - b^2 = (a-b)(a+b)\)

Why does this work? It's because if we expand \((a-b)(a+b)\), the middle terms cancel out, leaving only \(a^2 - b^2\).
In our exercise, we encounter this identity in the modified expression \((x-3)^2 - (2y)^2\). Here:

• \(a = x-3\)
• \(b = 2y\)

By applying the difference of squares rule, we get:

• \((x-3 - 2y)(x-3 + 2y)\)

In simple terms, anytime you can rewrite an expression in the form of one square term minus another, you can use this straightforward approach to simplify the equation.

Perfect Square Trinomials Demystified

A perfect square trinomial is a specific form of trinomial that can be expressed as the square of a binomial. That means, it perfectly factors into a repeated binomial term, exemplified by expressions like \((a+b)^2\) or \((a-b)^2\) .
The general form of a perfect square trinomial can be represented as:

• \(a^2 + 2ab + b^2 = (a+b)^2\)
• \(a^2 - 2ab + b^2 = (a-b)^2\)

In our context, \(x^2 - 6x + 9\) is a classic perfect square trinomial. Here's why:

• It follows the pattern \(a^2 - 2ab + b^2\), where \(a=x\) and \(b=3\).

As such, we can simplify it to \((x-3)^2\). The key to spotting perfect square trinomials lies in observing whether the first and last terms are squares and whether the middle term is twice the product of the square roots of the first and last terms. This recognition can make solving algebraic factorizations much less daunting.