The trick to this problem is to realize that #sqrt(xy)# can be simplified if either #x# or #y# is a perfect square. So, if we have #sqrt(16y)#, and we know that the square root of 16 is 4, we can rewrite this as #4sqrty#. Do you see? We need to do some factoring to get rid of all of the perfect squares under the #sqrt# sign, until we're left with a number that isn't a perfect square.

So, for the problem #sqrt162#, can we remove any perfect squares? If we think for a moment, we can see that #162 = 81 * 2#, and 81 is a perfect square (#9x9=81#). So this can be rewritten as

#sqrt((81)*(2))#

We can take the square root of 81 to get:

#9sqrt(2)#

which is as simplified as we can get in this case.