To rewrite the irrational square root \( \sqrt{48} \) as the product of an integer and another square root, we can first factor \( 48 \) into its prime factors:
\[ 48 = 16 \times 3 \]
Now, we can take the square root of \( 16 \):
\[ \sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3} \]
Therefore, \( \sqrt{48} \) can be rewritten as:
\[ \sqrt{48} = 4\sqrt{3} \]