Problem 90 - Beginning and Intermediate Algebra

Question

Find the following special products. $$ \left(\frac{4}{3}+z\right)\left(\frac{4}{3}-z\right) $$

Step-by-Step Solution

Answer

The short answer to the problem is: $$ \left(\frac{4}{3}+z\right)\left(\frac{4}{3}-z\right) = \frac{16}{9} - z^2 $$

Step 1: Identify the values of a and b

In this exercise, we have: $$ \left(\frac{4}{3}+z\right)\left(\frac{4}{3}-z\right) $$ The first term in both parentheses is \(\frac{4}{3}\), so \(a=\frac{4}{3}\), and the second term in both parentheses is \(z\), so \(b=z\).

Step 2: Apply the difference of squares formula

Now, apply the difference of squares formula: $$ (a+b)(a-b) = a^2 - b^2 $$ Substitute \(\frac{4}{3}\) for \(a\) and \(z\) for \(b\): $$ \left(\frac{4}{3}+z\right)\left(\frac{4}{3}-z\right) = \left(\frac{4}{3}\right)^2 - z^2 $$

Step 3: Calculate the squares

Calculate the square of \(\frac{4}{3}\) and the square of \(z\): $$ \left(\frac{4}{3}\right)^2 = \frac{16}{9} $$ and $$ z^2 = z \cdot z $$

Step 4: Substitute the squares back into the expression

Now, substitute the squares back into the expression: $$ \left(\frac{4}{3}\right)^2 - z^2 = \frac{16}{9} - z^2 $$ The product of the given binomials is: $$ \left(\frac{4}{3}+z\right)\left(\frac{4}{3}-z\right) = \frac{16}{9} - z^2 $$

Key Concepts

Understanding Special Products

Special products in algebra are expressions that simplify into more manageable forms using specific formulas. One of the most common types is the "difference of squares" formula. It applies to expressions like

• \((a+b)(a-b)\)

where it can be simplified to:

• \(a^2 - b^2\)

This formula is pivotal because it allows you to transform a multiplication problem into a simple subtraction of squares, making it easier to handle. In the example problem, by recognizing the pattern, you directly use this formula to find the solution without expanding everything. Here, \(a = \frac{4}{3}\) and \(b = z\), allowing simplification into:

• \(\left(\frac{4}{3}\right)^2 - z^2\)

Thus, understanding these shortcuts in algebra, can save time and reduce errors.

What are Binomials?

A binomial is an algebraic expression containing exactly two terms. These terms can be numbers, variables, or both. Take, for instance, \(\left(\frac{4}{3}+z\right)\).This expression consists of the constant term \(\frac{4}{3}\) and the variable \(z\).The beauty of binomials lies in their simplicity and the various ways they can be manipulated through algebraic operations.
When you multiply two conjugate binomials, like \((a+b)(a-b)\), it results in the difference of the squares of the two terms \(a\) and \(b\).
This leads to efficient simplification processes as shown in the original problem, turning a potentially complex task into a straightforward one.

Basic Algebra Concepts Involved

Algebra is all about finding and describing patterns. In this exercise, we utilized several core algebra concepts:

• **Identifying Patterns:** Recognizing the structure \((a+b)(a-b)\) as a difference of squares helps swiftly simplify the problem.
• **Substitution:** Replacing variables with numbers or other expressions is a crucial step for simplification.
• **Squaring Numbers:** Calculating \(\left(\frac{4}{3}\right)^2\) by multiplying \(\frac{4}{3}\times \frac{4}{3}=\frac{16}{9}\) illustrates how to handle fractions in algebraic operations.

These concepts are fundamental in algebra and assist in solving various mathematical problems. With practice, students will find such operations becoming second nature, allowing them to tackle even more complicated tasks with ease.