Problem 31 - Intermediate Algebra

Question

Simplify. See Examples 3 and 4 $$ \sqrt{32} $$

Step-by-Step Solution

Answer

The simplified form of \(\sqrt{32}\) is \(4\sqrt{2}\).

Step 1: Identify Prime Factorization

First, identify the prime factorization of 32. To do this, divide 32 by the smallest prime number possible. Begin with 2, the smallest prime number: 32 divided by 2 is 16. 16 divided by 2 is 8. 8 divided by 2 is 4. 4 divided by 2 is 2. Finally, 2 divided by 2 is 1. Thus, the prime factorization of 32 is \(2^5\).

Step 2: Apply the Square Root Property

Apply the square root property \(\sqrt{a^2} = a\) to the prime factors of 32. When there are pairs of the same factor, they can be taken out of the square root. In \(32 = 2^5\), there are two complete pairs of the number 2 since \(5 \div 2 = 2\) with a remainder of 1. Each pair of 2 can be taken out of the square root, while the single unpaired factor remains inside.

Step 3: Simplify the Expression

Since \(2^2 = 4\) and \(\sqrt{4} = 2\), we can simplify part by part. Thus, \(\sqrt{32} = \sqrt{2^5} = \sqrt{(2^2) imes (2^2) imes 2} = 2 \times 2 \times \sqrt{2} = 4\sqrt{2}\).

Key Concepts

Prime Factorization

Prime factorization is a method used to break down a number into its prime number components. This step is crucial when simplifying square roots because it helps identify which numbers can pair up and move outside the square root. For our example with the number 32, we start by dividing it by the smallest prime number, which is 2.

• Divide 32 by 2 to get 16.
• Divide 16 by 2 to get 8.
• Divide 8 by 2 to get 4.
• Divide 4 by 2 to get 2.
• Finally, divide 2 by 2 to get 1.

By dividing continuously until you reach 1, you determine that 32 is made up of five 2s, or expressed mathematically as \(2^5\). Recognizing this allows for easier manipulation in subsequent steps.

Square Root Property

The square root property is a helpful tool in algebra, stating that if you have a square root of a square number, it simplifies to the base number. Essentially, it means \(\sqrt{a^2} = a\). This property is incredibly powerful when you're dealing with prime factors because it enables you to simplify square roots by grouping.
For \(\sqrt{32}\), using the prime factorization \(2^5\), you want to identify pairs because each pair of identical numbers under a square root can 'escape' from the square root in one copy of the number.

• In \(2^5\), we find two complete pairs of 2s since \(5 \div 2 = 2\) with one 2 left over.
• Each of those pairs \((2^2)\) can become just \(2\) again outside the square root.

This helps reduce the complexity of the expression by moving numbers out from under the square root sign.

Intermediate Algebra

Intermediate Algebra involves solving more complex problems than basic algebra, often involving roots and factorization. The process of simplifying \(\sqrt{32}\) is a common type of problem in this area of algebra. Here, we bring everything together.
By following the rules of prime factorization and square root properties:

• We had \(\sqrt{32} = \sqrt{2^5}\).
• We simplified this by removing pairs: \(\sqrt{(2^2) \times (2^2) \times 2}\).
• Each \((2^2)\) came out of the square root as \(2\), and we multiplied them to get \(4\). What remains inside is \(\sqrt{2}\).

Thus, the final simplified form is \(4\sqrt{2}\). Understanding these steps makes handling root simplifications and other Intermediate Algebra challenges much easier.