Problem 80 - Algebra A Combined Function

Question

Mixed Practice Multiply. $$ (6 r-2 x)(6 r+2 x) $$

Step-by-Step Solution

Answer

The expression simplifies to \(36r^2 - 4x^2\).

Step 1: Recognize the Pattern

The expression \((6r - 2x)(6r + 2x)\) is a difference of squares pattern. Recall that for any expressions \(a\) and \(b\), \((a - b)(a + b) = a^2 - b^2\).

Step 2: Identify \(a\) and \(b\)

In \((6r - 2x)(6r + 2x)\), identify \(a = 6r\) and \(b = 2x\). These correspond to the terms that will be squared in the difference of squares pattern.

Step 3: Apply the Difference of Squares Formula

Substitute \(a = 6r\) and \(b = 2x\) into the formula \(a^2 - b^2\). Thus, the expression simplifies to \((6r)^2 - (2x)^2\).

Step 4: Calculate \(a^2\) and \(b^2\)

First, calculate \(a^2 = (6r)^2 = 36r^2\). Next, calculate \(b^2 = (2x)^2 = 4x^2\).

Step 5: Subtract the Squares

Substitute these values into the difference of squares formula: \(a^2 - b^2 = 36r^2 - 4x^2\). Therefore, the product of \((6r - 2x)(6r + 2x)\) is \(36r^2 - 4x^2\).

Key Concepts

Algebraic Expressions

In algebra, we often work with expressions that include variables and constants. These are known as algebraic expressions. They could be as simple as a single variable, like \(x\), or more complex, including terms that are added, subtracted, multiplied, or divided. For example, in our exercise, the expression \((6r - 2x)(6r + 2x)\) includes two terms in each parenthesis. Here, both "6r" and "2x" are terms within the algebraic expression.

Each term in an algebraic expression can consist of numbers (coefficients), variables, and even exponents. In our exercise, "6" is the coefficient of \(r\), and "2" is the coefficient of \(x\). These coefficients show us how many times these variables are counted. With this knowledge, understanding how to manipulate and transform these expressions becomes vital in solving algebraic problems.

Factoring Techniques

Factoring is a method used in algebra to simplify expressions and solve equations. It involves breaking down a complex expression into simpler components that, when multiplied together, give the original expression. One fundamental factoring technique is recognizing patterns, such as the difference of squares.

One special pattern highlighted in our exercise is the difference of squares, which is expressed as \((a - b)(a + b) = a^2 - b^2\). This pattern is useful when the expression appears as the product of a sum and a difference. In our example, factoring \((6r - 2x)(6r + 2x)\) is done by treating it as \((a - b)(a + b)\), where \(a = 6r\) and \(b = 2x\). By recognizing these as our "a" and "b," we can then apply the formula to simplify it into \(a^2 - b^2 = 36r^2 - 4x^2\).

Mastering such factoring techniques allows us to handle even more complicated expressions in algebra quickly and efficiently.

Polynomial Multiplication

Multiplying polynomials requires using properties of multiplication and understanding how to distribute terms correctly. The distributive property plays a crucial role here, allowing us to expand expressions methodically. In our exercise, we looked at a type of multiplication that uses the difference of squares formula.

Normally, to multiply polynomial expressions like \((6r - 2x)\) and \((6r + 2x)\), you would distribute each term in the first set of parentheses to each term in the second. However, because of the difference of squares pattern, we apply a shortcut: \((a - b)(a + b) = a^2 - b^2\). This reduces the work because we don’t have to multiply each term individually as we would in other cases.

In our example, knowing the "a" and "b" values, this shortcut swiftly transforms \((6r - 2x)(6r + 2x)\) into \(36r^2 - 4x^2\). This method helps us tackle polynomial multiplication efficiently and reinforces our understanding of significant algebraic patterns.