Take a factor out of the square root:

Question

√48x^2, where x≤0

Answer 1

x comes out, as x^2 = x^2

48 is 16 x 3, so 4 comes out 16 is a perfect square. 4^2 = 16

leaving only 3 inside the radical

4xsqr3 = sqr(48x^2)

the constraint x is less than or equal to zero doesn't seem to make any difference

when you square x it's the same value whether x is negative or positive

factor out an x will make 4xsqr3 negative

so add in 4xsqr3 < 0 or = 0

Answer 2

First you need to find perfect square factors of 48

Check some factors of 48;

2*24. nope

3*16 yes because 16 is perfect square and 3 is prime so

√48 = √16*√3 = 4√3

we had also √x^2 = x

answer is 4x√3