Problem 28 - Elementary and Intermediate Algebra

Question

Simplify each radical expression. All variables represent positive real numbers. $$ \sqrt[5]{64 x^{10} y^{5}} $$

Step-by-Step Solution

Answer

Simplified form: \( 2 x^2 y \sqrt[5]{2} \).

Step 1: Break Down the Radical Expression

The given expression is \( \sqrt[5]{64 x^{10} y^{5}} \). This is a fifth root of a product. Begin by factoring each part of the radicand separately. First, notice that \( 64 = 2^6 \), so the base of the expression can be rewritten using its prime factorization.

Step 2: Re-express the Expression with Fractional Exponents

We can rewrite the expression using fractional exponents: \( (2^6 x^{10} y^{5})^{1/5} \). The next step involves distributing the exponent into each part of the product: \( (2^6)^{1/5} \cdot (x^{10})^{1/5} \cdot (y^{5})^{1/5} \). This separates the fifth root across the multiplicands.

Step 3: Simplify Each Factor

Simplify each component. For \( (2^6)^{1/5} \), simplify to \( 2^{6/5} \), which can be further simplified. For \( (x^{10})^{1/5} \), the exponents multiply to become \( x^{10/5} = x^2 \). For \( (y^{5})^{1/5} \), the exponents multiply to become \( y^{5/5} = y \).

Step 4: Simplify Further if Possible

After simplification, your expression should now look like \( 2^{6/5} \cdot x^2 \cdot y \). Notice that \( 2^{6/5} = (2^5 \cdot 2)^{1/5} \), which simplifies to \( 2 \cdot 2^{1/5} \). Substitute this back into the expression for a simplified form: \( 2 x^2 y \cdot 2^{1/5} \).

Step 5: Consolidate the Expression

Combine the constants to simplify: \( 2 x^2 y \sqrt[5]{2} \). This is the fully simplified form of the original expression.

Key Concepts

Fractional Exponents

Fractional exponents provide a powerful way to express roots and powers in a single, unified form. They allow for easier manipulation during calculations, especially when simplifying radical expressions.

When you see a radical, such as a square root or fifth root, it can be rewritten using fractional exponents. For instance, the fifth root of a number is equivalent to raising that number to the power of \( \frac{1}{5} \).

• So, when you have the expression \( \sqrt[5]{a} \), it can be rewritten as \( a^{1/5} \).
• This helps when simplifying, as you can apply the rules of exponents, such as power of a power, to easily manage the expression.

For example, in the expression \( (2^6 x^{10} y^{5})^{1/5} \), each part is simplified individually by multiplying the exponents:

• \( 2^6 \) becomes \( 2^{6/5} \).
• \( x^{10} \) becomes \( x^{10/5} \), which simplifies to \( x^2 \).
• \( y^5 \) becomes \( y^{5/5} \), simplifying to \( y \).

This streamlines the entire process and provides a clearer path to the final simplified expression.

Fifth Roots

Fifth roots are a particular type of radical expression where you're finding a number that, when multiplied by itself five times, equals the original number. This concept is crucial when you want to simplify expressions involving\( \sqrt[5]{...} \).

Taking the fifth root of a prime factorized number like \( \sqrt[5]{64} \) is made simpler by recognizing \( 64 \textbf{} 2^6 \).

To simplify:

• First, note that due to \( 64 \equiv 2^6 \), the fifth root can be found using fractional exponents as \( (2^6)^{1/5} \).
• Doing this gives \( 2^{6/5} \), where the fraction \( 6/5 \) represents both the power and root involved.
• This could further simplify to \( 2 \times 2^{1/5}, \) recognizing \( 2^{1} \) part and the remaining \( 2^{1/5} \).

Through understanding and using the properties of radicals and exponents, fifth roots and similar operations become more approachable.

Prime Factorization

Prime factorization is the method of breaking down a number into its basic building blocks - the prime numbers. This process is especially helpful when working with radicals, as it allows for easier root finding and simplification.

For example, to find the prime factors of 64, consider:

• Divide 64 by the smallest prime number (which is 2): \( 64 \div 2 = 32 \).
• Continue dividing by 2: \( 32 \div 2 = 16 \).
• Keep on dividing: \( 16 \div 2 = 8 \), then \( 8 \div 2 = 4 \), and finally \( 4 \div 2 = 2 \).
• Resulting in \( 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \), or \( 2^6 \).

With its prime factorization, a number like 64 becomes easier to handle when simplifying expressions like radicals. In such a setting:

• Knowing that \( 64 = 2^6 \) and recognizing the fifth root, \( (2^6)^{1/5} \), helps in rewiring the way we communicate with numbers.
• This fundamental tool provides a means to break down complex expressions into manageable pieces, making advanced algebra significantly more approachable.

Prime factorization is a foundational concept that opens up opportunities for greater comprehension and accurate simplification of radical expressions.