First, break up the number into it's prime factors and put everything under the radical.
\(\sqrt{48} = \sqrt{2*2*2*2*3}\)
Then find the pairs, pull them out of the radical, and multiply the radical by that number.
We have 2 pairs of "2s" in the prime factorization of 48, so after removing them, we get
\(2*2\sqrt{3}\) which simplifies to \(4\sqrt{3}\)
Or you can try and factor the original number into perfect squares, for example
\(\sqrt{48} = \sqrt{4*4*3}\) and since √4 is equal to 2, you can remove the √4s inside the radical and replace them with "2s" outside the radical. Both ways will ive you the same answer.
If this was not clear, (I understand if it isn't) speak up so somebody can clarify.
I hope you find this helpful! :)