Problem 56 - Elementary and Intermediate Algebra

Question

Factor. See Example 5 or Example 7. $$ 6 a^{2}+6 b^{2}-13 a b $$

Step-by-Step Solution

Answer

The factorization is \((3a-2b)(2a-3b)\).

Step 1: Identify the structure

The expression given is a quadratic in terms of two variables: \( a \) and \( b \). It's structured as \( 6a^2 + 6b^2 - 13ab \). Notice that this resembles a quadratic trinomial.

Step 2: Reorganize the expression

Rewriting the expression to emphasize its quadratic structure could help in factoring: \( 6a^2 - 13ab + 6b^2 \). This reorganization doesn't change the expression but highlights its similarity to the standard form \( ax^2 + bx + c \).

Step 3: Factor by trial or grouping

Consider potential binomial factors. Since the leading and constant coefficient is 6, equal or similar two terms should be paired across different groups. Here, we'll proceed with trying factors like \((3a - 2b)\) and \((2a - 3b)\) for their cross-multiplication which makes them simplify without using long division directly.

Step 4: Multiply and verify

Check the factorization: Multiply \((3a - 2b)(2a - 3b)\), which gives: \[ 3a(2a - 3b) - 2b(2a - 3b) = 6a^2 - 9ab - 4ab + 6b^2 = 6a^2 - 13ab + 6b^2 \]. This confirms that the expression is correctly factored.

Key Concepts

Understanding Quadratic Expressions

Quadratic expressions are algebraic expressions where the highest exponent of the variable is two. In our original exercise, we have two variables, \(a\) and \(b\). The expression is \(6a^2 + 6b^2 - 13ab\). This is called a trinomial because it has three terms, each involving a combination of \(a\) and \(b\). It's crucial to recognize the quadratic nature here because it guides us toward potential factoring strategies.

Quadratics generally have forms like \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are coefficients. But in our case, since there are two variables, it appears as \(6a^2 + 6b^2 - 13ab\). Here:

• \(6a^2\) and \(6b^2\) are the quadratic terms.
• \(-13ab\) is the linear combination of two variables.

Recognizing such patterns helps a lot in tackling quadratic expressions in algebraic problems.

Exploring Binomial Factors

To factor quadratic expressions like \(6a^2 + 6b^2 - 13ab\), we often look for binomial factors. Binomials are simply expressions with two terms. A classic example is \((x + y)\).

In our exercise, we aim to rewrite the trinomial in terms of two binomials. This process may involve a bit of trial and error to find which pairs multiply to give the original quadratic expression. For the provided solution, the binomials \((3a - 2b)\) and \((2a - 3b)\) were chosen based on educated guesses and strategic pairing.

Why do binomials matter? They allow us to break down complex expressions into simpler components. Once we determine the correct binomials, we verify their multiplication to ensure they match the original expression. This step confirms if our choice of binomials was accurate or not.

Applying Algebraic Factorization

Algebraic factorization is the technique used to express an arithmetic expression as a product of its factors. In factoring polynomials, especially quadratic ones, the goal is to find factor expressions whose product is the original polynomial.

This involves identifying patterns, using methods like grouping, and testing potential factors. For the given exercise, factorization was achieved using the binomials \((3a - 2b)\) and \((2a - 3b)\).

We verify our factorization by expanding \((3a - 2b)(2a - 3b)\). This involves distributing each term in the first binomial by each term in the second, adding the results:

• Multiply \(3a\) by \(2a\) to get \(6a^2\).
• Multiply \(3a\) by \(-3b\) to get \(-9ab\).
• Multiply \(-2b\) by \(2a\) to get \(-4ab\).
• Multiply \(-2b\) by \(-3b\) to get \(6b^2\).

Assure that the terms combine to form the original expression \(6a^2 - 13ab + 6b^2\). This confirmation establishes that the factorization is indeed correct. Understanding and applying algebraic factorization serves as a foundational skill that enables students to solve more complex algebraic problems with confidence.