<Question>
Simplify $\sqrt{242}$.
</Question>

<think>
<Plan_1>
Okay, so I need to simplify the square root of 242. Hmm, let me think about how to approach this. I remember from math class that simplifying a square root involves factoring the number into perfect squares and other factors. The goal is to take out the square factors from under the radical sign. Let me try to recall the steps.

First, I should factor 242 into its prime factors. Let me start by checking if 242 is divisible by smaller prime numbers. Let's see, 242 is an even number because it ends with a 2, so it's divisible by 2.

<prompt_1.1> Factor 242 into its prime factors as per the above plan. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Dividing 242 by 2 gives me 121. Oh, wait, 121 is a perfect square, right? Because 11 times 11 is 121. So, that means 242 can be factored into 2 times 121, and 121 is 11 squared. So, putting that together, 242 = 2 × 11².
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, using the property of square roots that says √(a × b) = √a × √b, I can separate the square root of 242 into the product of the square roots of 2 and 11².
So, √(2 × 11²) = √(11²) × √2. Since the square root of 11 squared is just 11, that simplifies to 11 × √2.

Let me check if that's correct. If I square 11√2, I should get back to 242.

<prompt_2.1> Verify the simplification by squaring 11√2 and checking if it equals 242. </prompt_2.1>

Is there any other perfect square factor in 242? Let me double-check the prime factorization.

<prompt_2.2> Double-check the prime factorization of 242 and confirm the simplification of √242. </prompt_2.2>

Wait, just to make sure I didn't miss any other factors. Let me try dividing 242 by other primes.

<prompt_2.3> Verify the prime factorization of 242 and confirm the simplification of √242 by checking divisibility by other primes. </prompt_2.3>

Another way to check is to use the method of prime factorization step by step. Let's break down 242:

<prompt_2.4> Use step-by-step prime factorization to confirm the simplification of √242. </prompt_2.4>

Alternatively, if I use a calculator to approximate both √242 and 11√2, they should be equal. Let's see, √242 is approximately 15.556.
And 11√2 is approximately 11 × 1.4142 ≈ 15.556. So that matches.
</Plan_2>

<Execution_2>
<execution_2.1>
(11√2)² = 11² × (√2)² = 121 × 2 = 242. Yes, that works out. So, the simplified form of √242 is 11√2.
</execution_2.1>

<execution_2.2>
We had 242 divided by 2 is 121, and 121 is 11². There are no other factors. 11 is a prime number, so we can't factor it further. Therefore, the only perfect square factor in 242 is 121, which is 11 squared. So, the simplification should indeed be 11√2.
</execution_2.2>

<execution_2.3>
After dividing by 2, we got 121, which is 11 squared. So the prime factors are 2 and 11. Since 11 is a prime number, we can't factor it further. Therefore, the only perfect square factor in 242 is 121, which is 11 squared. So there are no other perfect squares in the factorization.
</execution_2.3>

<execution_2.4>
Start with 242. Divide by 2: 242 ÷ 2 = 121. Then 121 divided by 2 is 60.5, which isn't an integer, so 2 is only a factor once. Then 121 is 11 × 11. So yes, the prime factors are 2, 11, and 11. So 2 × 11². Therefore, the square factors are 11², which we can take out of the square root. So, yes, √242 simplifies to 11√2.
</execution_2.4>
</Execution_2>

<Final_answer>
\boxed{11\sqrt{2}}

\]
</Final_answer>
</think>