Problem 24 - Algebra 2

Question

Simplify. Assume that no variable equals 0. $$ \left(-2 r^{2} s\right)^{3}\left(3 r s^{2}\right) $$

Step-by-Step Solution

Answer

The simplified expression is \(-24r^7s^5\).

Step 1: Simplify the Exponent

First, simplify the term \((-2r^2s)^3\). To do this, apply the power to each element inside the parentheses:\((-2)^3 = -8,(r^2)^3 = r^{6},(s)^3 = s^{3}\) Therefore, we have \((-2r^2s)^3 = -8r^6s^3\).

Step 2: Multiply the Simplified Term with the Second Term

Now, we multiply \(-8r^6s^3\) with the second term, \((3rs^2)\):\(-8r^6s^3 imes 3rs^2 = (-8 imes 3) imes (r^6 imes r) imes (s^3 imes s^2)\).Simplifying each part, we get:- \(-8 imes 3 = -24\) - \(r^6 imes r = r^{7}\) - \(s^3 imes s^2 = s^{5}\)

Step 3: Write the Final Expression

Combine all the simplified parts to write the final expression:\[-24r^7s^5\]

Key Concepts

Exponentiation

When dealing with algebraic expressions, exponentiation is a core concept that involves dealing with powers of numbers or variables. In this case, you are given an expression \((-2r^2s)^3\). This means each element inside the parentheses is raised to the power of three. Here’s how you break it down:

• Raise \(-2\) to the power of 3, which results in \(-8\).
• Raise \(r^2\) to the power of 3. This involves multiplying the exponent 2 by 3 to get \(r^6\).
• Raise \(s\) to the power of 3, giving you \(s^3\).

Each of these steps applies the rule that when raising a power to another power, you multiply the exponents. After simplifying, \((-2r^2s)^3 = -8r^6s^3\). This makes it easier to further simplify and perform multiplication with other terms.

Polynomial Multiplication

Polynomial multiplication involves multiplying expressions with several terms. In this exercise, after simplifying the exponentiated term, \(-8r^6s^3\), you need to multiply it by another polynomial, \(3rs^2\). Here are the steps:

• Multiply the numerical coefficients: \(-8 \times 3 = -24\).
• Multiply the variables with the same base by adding their exponents:
  • For \(r\), \(r^6 \times r^1 = r^{6+1} = r^7\).
  • For \(s\), \(s^3 \times s^2 = s^{3+2} = s^5\).

This results in the expression \(-24r^7s^5\). Such multiplication is essential when dealing with algebraic expressions as it combines different terms efficiently while obeying the mathematical rules for powers.

Simplifying Expressions

Simplifying expressions means reducing them to their simplest form, making them easier to work with. In this exercise, following the completion of the exponentiation and multiplication, you're left with the task of combining everything into a single expression: \(-24r^7s^5\).
Simplifying ensures that the expression is tidy and concise. Here’s the simplified form, ensuring:

• All terms are fully combined.
• Correct application of arithmetic operations.
• Exponents are correctly calculated and presented.

This does not just aid in clarity but also improves the handling of expressions for further mathematical operations or evaluations. It is crucial for accuracy and efficiency in algebra, setting up the groundwork for more complex problem-solving scenarios.

Variables in Algebra

Variables are symbols used to represent numbers in algebra. They allow generalization of rules and operations, making it possible to solve equations and perform operations across different categories of numbers.
In the provided exercise, variables \(r\) and \(s\) play a crucial role:

• They denote unknown quantities that can be manipulated through algebraic operations.
• They carry exponents, which interact during multiplication to form new expressions like \(r^7\) and \(s^5\).
• These interactions demonstrate the properties of exponents and the foundational operations in algebra.

Understanding how to work with variables is foundational in algebra. It not only aids students in visualizing the problem but also in comprehending the systematic approach to mathematical expressions and the application of algebraic rules.