Problem 90 - Intermediate Algebra

Question

Perform the multiplications. All variables represent positive real numbers. See Example \(9 .\) $$ y^{2 / 5}\left(y^{-2 / 5}+y^{3 / 5}\right) $$

Step-by-Step Solution

Answer

The expression simplifies to \( 1 + y \).

Step 1: Distribute the Term

First, distribute the term \( y^{2/5} \) across each term inside the parentheses: \( y^{2/5} \times y^{-2/5} + y^{2/5} \times y^{3/5} \).

Step 2: Apply the Product of Powers Rule

According to the product of powers rule, we can add the exponents when multiplying like bases. For the first term: \( y^{2/5} \times y^{-2/5} = y^{(2/5) + (-2/5)} = y^0 \). For the second term: \( y^{2/5} \times y^{3/5} = y^{(2/5) + (3/5)} = y^{5/5} = y^1 \).

Step 3: Simplify the Expressions

Simplify each expression based on the exponents calculated: \( y^0 = 1 \) and \( y^1 = y \). Thus, the expression simplifies to \( 1 + y \).

Key Concepts

Understanding the Product of Powers Rule

When working with exponents, the product of powers rule is crucial. This rule states that when you multiply two expressions with the same base, you simply add their exponents. To illustrate, consider an expression like \( a^m \times a^n \). Here, both terms have the same base \( a \). According to the product of powers rule, the result is \( a^{m+n} \). In our original exercise:

• The first part is \( y^{2/5} \times y^{-2/5} \), resulting in \( y^{(2/5) + (-2/5)} = y^0 \).
• The second part is \( y^{2/5} \times y^{3/5} \), leading to \( y^{(2/5) + (3/5)} = y^{5/5} = y^1 \).

This method of adding exponents is what makes calculations involving powers simpler and more efficient. You just need to ensure the bases are the same.

Exploring the Distributive Property

The distributive property is a fundamental concept that helps in multiplying expressions more effectively. It is applicable when you have a term outside a bracket that needs to be multiplied by each term inside the bracket. For example, in our expression \( y^{2/5}(y^{-2/5} + y^{3/5}) \), you must distribute \( y^{2/5} \) to each term inside the parentheses. So, you perform two separate multiplications:

• \( y^{2/5} \times y^{-2/5} \)
• \( y^{2/5} \times y^{3/5} \)

This approach of distributing ensures that each term inside the parenthesis is correctly multiplied by the outer term. The distributive property is a powerful tool in algebra that simplifies complex polynomial expressions.

Working with Positive Exponents

Positive exponents indicate the number of times a base is used as a factor in multiplication. For instance, \( a^3 \) means \( a \times a \times a \). In our problem, we see exponents expressed as fractions and positive whole numbers. When you simplify expressions with positive exponents:

• \( y^1 \) simplifies to just \( y \), meaning the base is used one time.
• \( y^{5/5} = y^1 \), where the fraction simplifies to the whole number 1, retaining the same base.

Exponents make it easier to handle repeated multiplication. Always ensure the exponents remain positive in expressions for straightforward simplification.