Question 683835
<pre>
Only one kind of root is called a "SQUARE" root.  Sometimes students
get the wrong impression that all roots are called "SQUARE" roots, but
this is false.

simplify by factoring 5 square root 96x^12y^15

Did you mean this {{{root(5,96x^12y^15)}}}?

If so that is NOT a "SQUARE" root but a FIFTH root.

If it were as you stated it would be {{{5*sqrt(96x^12y^15)}}}

which is an entirely different problem.

If you meant the FIFTH root, not the SQUARE root, then you change
{{{x^12}}} to {{{x^10*x^2}}} because 10 is the largest exponent
divisible by 5.  We also change {{{96}}} to {{{32*3}}} or {{{2^5*3}}}.
We do not need to change {{{y^15}}} because 15 is already divisible
by 5, the index of the 5th root.

{{{root(5,96x^12y^15)}}} = {{{root(5,32*3*x^10*x^2*y^15)}}} =

{{{root(5,2^5*3*x^10*x^2*y^15)}}} =
  
Now we divide each exponent that is divisible by 5 and bring it out
from under the radical with the exponent divided by the index 5:

{{{3^1*x^2*y^3*root(5,3*x^2)}}} and you can drop the 1 exponent for x

{{{3*x^2*y^3*root(5,3*x^2)}}}

If you really meant "SQUARE ROOT", like this:

{{{5*sqrt(96x^12y^15)}}}

Then the index is understood to be 2.  I'll put the index:

{{{root(2,96x^12y^15)}}}

then you do not need to change {{{x^12}}} because 12 is already
divisible by 2.  We change {{{96}}} to {{{16*6}}} or {{{2^4*6}}}, so
that the exponent 4 is divisible by 2.
We change {{{y^15}}} to {{{y^14*y}}}.

{{{5*root(2,16*6*x^12y^14*y)}}}

{{{5*root(2,2^4*6*x^12y^14*y)}}}

Now we divide each exponent by 2 that is divisible by 2 and bring it out
from under the radical with the exponent divided by the index 2:

{{{5*2^2*x^6*y^7*root(2,6*y)}}} =

{{{5*4*x^6*y^7*root(2,6*y)}}}

and we drop the 2 index because it's understood when the root is a
SQUARE root and not some other root such as the FIFTH root.

{{{5*4x^6y^7*sqrt(6y)}}} =

{{{20x^6y^7*sqrt(6y)}}}

Edwin</pre>