Chapter 13: Problem 37
Simplify. $$3 \sqrt[5]{32 x y^{11}}$$
Short Answer
Expert verified
The simplified form of the expression is \(6y^2 \root{5}{x y}\).
Step by step solution
01
Identify the Perfect Fifth Powers
First, identify any numbers inside the radical that are perfect fifth powers. In this case, 32 can be expressed as \(2^5\). Also, factor out the highest power of y that is a multiple of 5, which is \(y^{10} = (y^2)^5\). The expression becomes \(3 \times 2 \times y^2 \times \root{5}{x y}\).
02
Simplify the Radical
Simplify the expression by multiplying the coefficients outside the radical and apply the fifth root to the perfect fifth powers inside the radical. Thus, \(3 \times 2 \times y^2\) becomes \(6y^2\) and the expression simplifies to \(6y^2 \root{5}{x y}\).
03
Rewrite the Simplified Expression
Combine all the simplified components to get the final simplified expression: \(6y^2 \root{5}{x y}\). This is the simplified form of the original expression.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Perfect Fifth Powers
When working with radical expressions, particularly roots higher than the common square or cube roots, understanding perfect fifth powers is essential. A perfect fifth power is a number that can be expressed as another number raised to the power of five, such as \( 32 = 2^5 \). In simplifying radicals, it's crucial to identify these perfect fifth powers because they can be taken out of the radical, making the expression easier to handle.
Here's a quick list of the first few perfect fifth powers to aid in recognition:
Here's a quick list of the first few perfect fifth powers to aid in recognition:
- \( 1^5 = 1 \)
- \( 2^5 = 32 \)
- \( 3^5 = 243 \)
- \( 4^5 = 1024 \)
- \( 5^5 = 3125 \)
Radical Expressions
Radical expressions involve roots, such as square roots \(\sqrt{}\), cube roots \(\sqrt[3]{}\), and in our case, fifth roots \(\sqrt[5]{}\) of numbers. These expressions can look intimidating, but they follow rules similar to regular arithmetic. The key to understanding radicals is realizing that they're the opposite operation of raising a number to a power. For instance, as \(2^5 = 32\), we know that \(\sqrt[5]{32} = 2\).
The complexity increases as variables get involved alongside constants inside the radical. In our example, the expression includes not just a constant but also the variables \(x\) and \((y^{11})\). Simplifying radical expressions often involves factoring out perfect powers, as shown in the provided solution, to reduce the radicals to their simplest form.
The complexity increases as variables get involved alongside constants inside the radical. In our example, the expression includes not just a constant but also the variables \(x\) and \((y^{11})\). Simplifying radical expressions often involves factoring out perfect powers, as shown in the provided solution, to reduce the radicals to their simplest form.
Simplification Process
The process of simplifying radicals is a systematic approach to making the expression more manageable. The idea is to remove as much from under the radical as possible by factoring out perfect powers. In our example, the fifth root of 32 is simplified by recognizing 32 as a perfect fifth power of 2, which can then be taken out of the radical.
To further simplify, one can also factor out the highest power of any variable that is a multiple of the index of the radical—in this case, 5. By reducing the exponent of \(y^{11}\) to \(y^{10} \times y\), \(y^{10}\) is factored out of the radical because it is a perfect fifth power of \(y^2\). The remaining \(y\), which is not a perfect power, stays within the radical.
Combining these simplified components outside of the radical yields the final expression, which is easier to use in further calculations or applications. Simplification can be applied in many areas, including geometry, algebra, and calculus, making it a vital skill in mathematics.
To further simplify, one can also factor out the highest power of any variable that is a multiple of the index of the radical—in this case, 5. By reducing the exponent of \(y^{11}\) to \(y^{10} \times y\), \(y^{10}\) is factored out of the radical because it is a perfect fifth power of \(y^2\). The remaining \(y\), which is not a perfect power, stays within the radical.
Combining these simplified components outside of the radical yields the final expression, which is easier to use in further calculations or applications. Simplification can be applied in many areas, including geometry, algebra, and calculus, making it a vital skill in mathematics.