To rewrite the irrational square root \( \sqrt{48} \) as the product of an integer and another irrational square root, we can break down \( 48 \) into its prime factors:
\[ 48 = 16 \times 3 \]
Since \( 16 \) is a perfect square, we can take it out of the square root:
\[ \sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3} \]
So the correct response is:
4 start root 3 end root (or \( 4\sqrt{3} \)).