Rewrite the irrational square root 48−−√ 48 as the product of an integer and another irrational square root.(1 point) Responses 224−−√ 2 24 2 start root 24 end root 212−−√ 2 12 2 start root 12 end root 43–√ 4 3 4 start root 3 end root 412−−√

To rewrite \( \sqrt{48} \) as the product of an integer and another irrational square root, we start by simplifying \( \sqrt{48} \).

First, factor \( 48 \) into its prime factors:
\[
48 = 16 \times 3 = 4^2 \times 3
\]

Now we can rewrite the square root:
\[
\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}
\]

So \( \sqrt{48} \) can be expressed as:
\[
\sqrt{48} = 4 \sqrt{3}
\]

The correct response is \( 4 \sqrt{3} \), represented as an integer \( 4 \) and the irrational square root \( \sqrt{3} \).

None of the options provided match this answer directly, but in the format you have shown, it would be best expressed as \( 4 \sqrt{3} \). Please verify against your options or recheck if they match or have similar representations.