Problem 100 - Intermediate Algebra

Question

Perform each indicated operation. See Sections 1.4 and 5.4 $$ (6 x)(8 x) $$

Step-by-Step Solution

Answer

The result is \(48x^2\).

Step 1: Understand the Problem

We need to multiply two expressions: \[(6x)(8x)\]This involves multiplying the coefficients (the numbers) as well as the variables, which in this case are both \(x\).

Step 2: Multiply the Coefficients

The coefficients in the expression are 6 and 8. First, multiply these two numbers together:\[6 \times 8 = 48\]

Step 3: Multiply the Variables

Both terms contain the same variable \(x\). When multiplying variables, you use the property of exponents that states:\[x^a \times x^b = x^{a+b}\]Since both terms are \(x^1\), when you multiply them, the exponents are added:\[x^1 \times x^1 = x^{1+1} = x^2\]

Step 4: Combine the Results

The results from the previous steps can be combined to give the final answer:\[(6x)(8x) = 48x^2\]

Key Concepts

Coefficients

In algebraic terms, coefficients are the numerical factor that multiplies a variable. They play a crucial role in algebraic expressions as they define the magnitude and direction of the variable. In our problem \((6x)(8x)\), the coefficients are the numbers 6 and 8.
But what do you do with coefficients when you're dealing with multiplication? The rule is simple: multiply them directly. In this case:

• Take 6 and 8, because these are the numbers directly in front of the \(x\).
• Multiply them: \(6 \times 8 = 48\).

Once you have the product of the coefficients, you are ready to pair it with the result from the variables. Coefficients essentially set the scale for the variable part of the term.

Exponents

Exponents indicate how many times a number, known as the base, is multiplied by itself. In algebra, when variables are involved, exponents can help simplify expressions and clarify how terms interact when operations like multiplication occur.
In the exercise \((6x)(8x)\), the variable part \(x\) has an implicit exponent of 1, often left unwritten because any number, including variables, raised to the power of 1 equals itself. When multiplying like bases, such as \(x\), you add their exponents:

• \(x^1 \times x^1\) becomes \(x^{1+1} = x^2\).

This property simplifies the multiplication process and gives a compact representation of the operations performed on the variables. It’s a key aspect not only in simplifying multiplication but in expanding expressions more generally.

Variables

Variables serve as placeholders for unknown numbers in equations or expressions, designated typically by letters such as \(x, y, z\). They allow equations to remain flexible and generalizable across different calculations and scenarios.
In the given exercise \((6x)(8x)\), \(x\) is the variable, representing a number not specified explicitly.

• Variables can have coefficients, which dictate their role in equations.
• When multiplied, like in our task, variables follow specific algebraic rules, such as combining exponents.

This ability to transform and adapt based on mathematical rules is what makes variables incredibly useful for both simple operations and complex algebraic manipulations. Understanding the flexibility and potential of variables is a foundational step in mastering algebra.