Question 134104
Perform a prime factorization of the quantity under the radical.  If any of the prime factors occur in pairs, remove the pair from under the radical and put a single instance of the factor outside the radical.  Once all the pairs have been removed, re-combine the remaining factors under the radical, and combine any factors you have moved outside of the radical.


Example:  {{{sqrt(97020)}}}


97020/2 = 48510  One factor of 2, still even
48510/2 = 24255  Another factor of 2, no longer even, but sum of digits div by 3
24255/3 = 8085   One factor of 3, sum of digits still div by 3
8085/3  = 2695   Another factor of 3, sum of digits not div by 3, but ends in 5
2695/5  = 539    One factor of 5, Does not end in 5, so try 7
539/7   = 77     One factor of 7
77/7    = 11     Another factor of 7
11               Is prime.


Factors:  2 * 2 * 3 * 3 * 5 * 7 * 7 * 11.


Pair of 2s, pair of 3s, and pair of 7s, so


{{{sqrt(97020)=2*3*7*sqrt(5*11)=42*sqrt(55)}}}.   And that is as simple as you can make it.


Example 2:


{{{sqrt(x^3-7x^2+16x-12)}}}


{{{x^3-7x^2+16x-12}}} divided by {{{x-3}}} = {{{x^2-4x+4}}} (use synthetic division to verify this for yourself).  And {{{x^2-4x+4}}} factors to {{{(x-2)(x-2)}}}, so:


{{{sqrt(x^3-7x^2+16x-12)=sqrt((x-2)^2(x-3))=(x-2)sqrt(x-3)}}}