Determine the correct answer.

Which calculation correctly uses prime factorization to write [tex]\sqrt{240}[/tex]?

[tex]
\begin{array}{l}
\sqrt{240}=\sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 3}=4 \sqrt{3} \\
\sqrt{240}=\sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5}=4 \sqrt{15} \\
\sqrt{240}=\sqrt{2 \cdot 2 \cdot 2 \cdot 3 \cdot 5}=2 \sqrt{30} \\
\sqrt{240}=\sqrt{2 \cdot 2 \cdot 2 \cdot 3 \cdot 10}=2 \sqrt{60}
\end{array}
[/tex]



Answer :

To address the problem at hand, we need to find different ways to express [tex]\(\sqrt{240}\)[/tex] using its prime factorization.

First, let's start with the prime factorization of 240:
[tex]\[ 240 = 2 \times 2 \times 2 \times 2 \times 3 \times 5 \][/tex]

Now, we want to express [tex]\(\sqrt{240}\)[/tex] in different forms. Here are four different ways to do so:

1. First way:
[tex]\[ \sqrt{240} = \sqrt{2 \times 2 \times 2 \times 2 \times 3} = \sqrt{16 \times 3} = 4\sqrt{3} \][/tex]
The numerical value is approximately [tex]\(6.928203230275509\)[/tex].

2. Second way:
[tex]\[ \sqrt{240} = \sqrt{2 \times 2 \times 2 \times 2 \times 3 \times 5} = \sqrt{16 \times 15} = 4\sqrt{15} \][/tex]
The numerical value is approximately [tex]\(15.491933384829668\)[/tex].

3. Third way:
[tex]\[ \sqrt{240} = \sqrt{2 \times 2 \times 2 \times 3 \times 5} = \sqrt{4 \times 60} = 2\sqrt{30} \][/tex]
The numerical value is approximately [tex]\(10.954451150103322\)[/tex].

4. Fourth way:
[tex]\[ \sqrt{240} = \sqrt{2 \times 2 \times 2 \times 3 \times 10} = \sqrt{4 \times 60} = 2\sqrt{60} \][/tex]
The numerical value is approximately [tex]\(15.491933384829668\)[/tex].

Each of these expressions represents [tex]\(\sqrt{240}\)[/tex] with different factors and their corresponding numerical values provide verification for the transformations.