Problem 13 - Elementary and Intermediate Algebra

Question

Fill in the blanks. $$ \begin{aligned} 5 \sqrt{8} \cdot 7 \sqrt{6} &=5(7) \sqrt{8} \\ &=35 \sqrt{\quad} \\ &=35 \sqrt{\quad\cdot3} \\ &=35(\quad) \sqrt{3} \\ &=140 \sqrt{3} \end{aligned} $$

Step-by-Step Solution

Answer

The answer is \( 140 \sqrt{3} \).

Step 1: Simplify Inside the Square Root

Notice that the expression involves the product of square roots: \( 5 \sqrt{8} \cdot 7 \sqrt{6} \). We begin by simplifying the product of the numbers inside the square roots, thus we multiply the square root parts: \( \sqrt{8} \cdot \sqrt{6} = \sqrt{8 \cdot 6} = \sqrt{48} \).

Step 2: Substitute Back the Simplified Expression

Now substitute \( \sqrt{48} \) back into the expression: \( 5 \cdot 7 \cdot \sqrt{48} = 35 \sqrt{48} \). This explains the partial expression seen in the problem, where it says \( 35 \sqrt{\quad} \). Fill in the blank with 48: \( 35 \sqrt{48} \).

Step 3: Factor and Simplify the Square Root

Next, we need to break down \( \sqrt{48} \) into its prime factors and simplify. The prime factors of 48 are \( 2^4 \cdot 3 \). Therefore, \( \sqrt{48} = \sqrt{2^4 \cdot 3} = \sqrt{16 \cdot 3} \). This simplifies to \( \sqrt{16} \cdot \sqrt{3} = 4 \sqrt{3} \).

Step 4: Final Substitution and Calculation

Substitute the simplified form \( 4 \sqrt{3} \) back into the expression: \( 35 \sqrt{48} = 35 \cdot 4 \sqrt{3} \). Calculate the multiplication: \( 35 \times 4 = 140 \). Therefore, the complete simplified answer is \( 140 \sqrt{3} \).

Key Concepts

Prime Factorization

In algebra, understanding prime factorization is crucial for simplifying square roots and radical expressions. To perform prime factorization, we break down a number into its prime number components.
For instance, in the original exercise where simplifying \( \sqrt{48} \) was necessary, we find its prime factors. These are the smallest prime numbers that, when multiplied together, give the original number.

Let's look at how this works:

• First, identify that 48 is an even number, so it can be divided by 2, \( 48 \div 2 = 24 \).
• Since 24 is also even, continue dividing by 2, \( 24 \div 2 = 12 \).
• Repeat, \( 12 \div 2 = 6 \).
• Again, divide by 2, \( 6 \div 2 = 3 \).
• Finally, 3 is a prime number, leaving us with the prime factorization of 48: \( 2^4 \cdot 3 \).

Performing prime factorization helps us to identify pairs of numbers, simplifying square roots more easily.

Multiplication of Radicals

When dealing with radicals, multiplying them can make simplification easier. Multiplication of radicals follows a simple rule: \( \sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b} \). This allows us to combine radicals under a single square root by multiplying their bases.
For example, in the exercise involving \( \sqrt{8} \cdot \sqrt{6} \), you multiply the numbers inside the radicals first: \( 8 \cdot 6 = 48 \).

This single radical, \( \sqrt{48} \), is many times easier to work with. Here's why:

• Instead of dealing with two separate square roots, consolidating them simplifies the calculation.
• Once multiplied, proceed with prime factorization to further simplify.

This method allows for a more structured simplification process, turning seemingly complex expressions into simpler forms.

Square Root Simplification

Simplifying square roots involves transforming expressions into their simplest form by extracting squares. After factoring through prime identification, you aim to simplify this expression by taking out pairs of prime numbers.
Let's consider the exercise's \( \sqrt{48} = \sqrt{2^4 \cdot 3} \). The foundation of such simplification is recognizing perfect squares involving the primes. Here’s how:

• The power of 4 on 2 suggests \( 2^2 \cdot 2^2 \), creating perfect squares \( 2 \cdot 2 = 4 \).
• Square roots negate these pairs: \( \sqrt{2^4} = \sqrt{(2^2)^2} = 2^2 = 4 \).
• The remaining factor 3 stays under the square root as \( \sqrt{3} \).
• Thus, \( \sqrt{48} = 4 \sqrt{3} \), a much more approachable form.

This process shows how organizing numerical factors makes handling radicals logical and efficient, always aiming for that "simplified" sweet spot.