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1. Write the prime factorization of the radicand. What is the simplified form of [tex]\( 3 \sqrt{135} \)[/tex]?

2. Apply the product property of square roots. Write the [tex]\( \sqrt{15} \)[/tex] radicand as a product, forming as many perfect square roots as possible.

[tex]\[ 3 \sqrt{5 \cdot 3} = 3 \sqrt{15} \][/tex]

3. Simplify.

[tex]\[ (3+3) \sqrt{5 \cdot 3} = 6 \sqrt{15} \][/tex]

[tex]\[ 3 \cdot 3 \sqrt{5 \cdot 3} = 9 \sqrt{15} \][/tex]


Sagot :

Certainly! Let's break down the mathematical steps to simplify [tex]\(3 \sqrt{135}\)[/tex] and solve this problem.

### Step 1: Prime Factorization of the Radicand
We start with the radicand, which is 135.

1. Prime Factorization:
- 135 is divisible by 3:
[tex]\[ 135 \div 3 = 45 \][/tex]
- 45 is also divisible by 3:
[tex]\[ 45 \div 3 = 15 \][/tex]
- 15 is again divisible by 3:
[tex]\[ 15 \div 3 = 5 \][/tex]
- 5 is a prime number.
- So, the prime factorization of 135 is:
[tex]\[ 135 = 3 \times 3 \times 3 \times 5 \][/tex]

### Step 2: Simplified Form of [tex]\(3 \sqrt{135}\)[/tex]
Next, simplify [tex]\(3 \sqrt{135}\)[/tex] using the identified prime factors.

1. Group the factors to form perfect squares:
[tex]\[ 135 = 3^2 \times 3 \times 5 = (3^2) \times (3 \times 5) \][/tex]

2. Apply the product property of square roots:
[tex]\[ \sqrt{135} = \sqrt{3^2 \times 3 \times 5} = \sqrt{3^2} \times \sqrt{3 \times 5} = 3 \times \sqrt{15} \][/tex]

3. Therefore:
[tex]\[ 3 \sqrt{135} = 3 \times (3 \times \sqrt{15}) = 9 \sqrt{15} \][/tex]

### Complete Simplification:
Putting it all together:

1. We have the initial form:
[tex]\[ 3 \sqrt{135} \][/tex]

2. Substituting the simplified square root form:
[tex]\[ 3 \sqrt{135} = 3 \times 3 \sqrt{15} = 9 \sqrt{15} \][/tex]

### Summary
- The prime factors of 135 are [tex]\([3, 3, 3, 5]\)[/tex].
- The initial prime factorization of 135 is [tex]\([3, 3, 3, 5]\)[/tex].
- The simplified expression for [tex]\(\sqrt{135}\)[/tex] is [tex]\(3 \sqrt{15}\)[/tex].
- Finally, [tex]\(3 \sqrt{135}\)[/tex] simplifies to [tex]\(9 \sqrt{15}\)[/tex].