Problem 63 - Elementary Algebra

Question

For the following problems, find the products. $$ \left(2 b+\frac{6}{7}\right)\left(2 b-\frac{6}{7}\right) $$

Step-by-Step Solution

Answer

Question: Calculate the product of the following expressions: \((2b + \frac{6}{7})(2b-\frac{6}{7})\). Answer: The product of the given expressions is \(4b^2 - \frac{36}{49}\).

Step 1: Identify the Difference of Squares Formula

Recognize that this multiplication problem is a difference of squares \((a+b)(a-b) = a^2 - b^2\). In this case, \(a = 2b\) and \(b = \frac{6}{7}\).

Step 2: Apply the Difference of Squares Formula

Using the formula and the values for \(a\) and \(b\), we can calculate the product as follows: $$ (2b + \frac{6}{7})(2b-\frac{6}{7}) = (2b)^2 - \left(\frac{6}{7}\right)^2 $$

Step 3: Calculate the Square of Each Term

Square the terms inside the parentheses: $$ (2b)^2 = 4b^2, \quad \left(\frac{6}{7}\right)^2 = \frac{36}{49} $$

Step 4: Subtract the Squared Terms to Find the Product

Now, subtract the squared terms to find the product: $$ 4b^2 - \frac{36}{49} $$

Key Concepts

Polynomial Multiplication

Polynomial multiplication is an essential part of algebra. It involves multiplying terms within polynomials to get a new polynomial. In simpler terms, it's a way to expand expressions. For example, when multiplying two binomials, such as \((a+b)(a-b)\), we leverage distributive properties to combine all parts together effectively.

The distributive property tells us that we need to multiply each term in the first polynomial by every term in the second. For two terms, each from two polynomials, this results in four individual multiplications typically. However, recognizing patterns like the difference of squares can simplify this process, making it possible to multiply certain forms with fewer steps.

Using special identities like the difference of squares \((a+b)(a-b) = a^2 - b^2\) helps us get to the solution more directly, skipping the intermediate steps of traditional polynomial multiplication. This specific pattern allows us to quickly identify similar tasks during homework.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations. They form the foundation of algebra, representing real-world scenarios symbolically. Understanding how to manipulate these expressions is key to solving mathematical problems efficiently.

The given problem is an example of an algebraic expression where we have to identify specific characteristics to simplify it. By seeing \((2b + \frac{6}{7})(2b - \frac{6}{7})\) we glimpse a recognizable pattern — the difference of squares. Furthermore, expressions like this open the door to more advanced math topics, allowing students to build strong skills.

• Variables: Here, \(b\) is a variable signifying an unknown quantity.
• Constants: Numbers like \(\frac{6}{7}\) remain unchanged during calculations.
• Operations: Include addition, subtraction, multiplication, or division, which transform these expressions.

Identifying and playing with these components is vital to mastering algebra.

Factoring

Factoring is the process of breaking down algebraic expressions into simpler components or 'factors' that, when multiplied, produce the original expression. It's like unwrapping the layers of an algebraic gift to understand what's inside.

In algebra, recognizing the difference of squares is a form of factoring where we see something that seems complex and simplify it in our minds and, on paper, manageably. To break down a difference of squares, \((a+b)(a-b) = a^2 - b^2\), we focus on the squared terms and their positions in subtraction.

Identifying these patterns is a powerful tool. As with our problem's form, applying this pattern means you can go straight to the solution rather than step by step multiplying and gathering similar terms diligently. Factoring does not just stop at a difference of squares. It's valuable for tackling quadratics and polynomials of higher degrees as well. This practice solidifies understanding of expressions, leading to confident algebraic problem-solving.