Square root of 6x multiplied by the square root of x+3

When you multiply 2 "square root expressions", you can put the multiplication problem under 1 square root. So,

√(6x) * √(3x2) = √(6x)*(3x2)

Follow the normal rules to multiply the expression under the square root,

√(18x3) 

Now put this expression in simple radical form, if required, which means factoring out all the "perfect squares" from under the square root.

Factor 18 into 9*2 because 9 is a perfect square (3*3)

Factor x3 into x2 * x because x2 is a perfect square

√18x3 = √(9x2 * 2x) = √9x2 * √2x

√9x2 = 3x   (because 3*3=9, x*x = x2, so 3x * 3x = 9x2)

√9x2 * √2x = 3x * √2x, which can also be written 3x√2x

idk

step-by-step explanation: i just don't