Question 314204
<pre><b>
The other tutor's solution is incorrect. He must have mistook
250 for 2500.

Later: He came back and fixed it!

{{{sqrt(250x^5y^3)}}}

250 is not a perfect square but it can be broken up into a perfect square
times a non-perfect square, for it can be broken up into 25*10, where 25 
is a perfect square and 10 is a non-perfect square. so write 250 as 25*10.

{{{sqrt(red(25)*10x^5y^3)}}}

{{{x^5}}} is not a perfect square but it can be broken up into a perfect square
times a non-perfect square, for it can be broken up into {{{x^4*x}}}, where {{{x^4}}}
is a perfect square and {{{x}}} is a non-perfect square. so write {{{x^5}}} as {{{x^4*x}}}.

{{{sqrt(red(25)*10red(x^4)*x*y^3)}}}

{{{y^3}}} is not a perfect square but it can be broken up into a perfect square
times a non-perfect square, for it can be broken up into {{{y^2*y}}}, where {{{y^2}}}
is a perfect square and {{{y}}} is a non-perfect square. so write {{{y^3}}} as {{{y^2*y}}}.

{{{sqrt(red(25)*10red(x^4)*x*red(y^2)*y)}}}

Notice that I've colored all the perfect squares red. We can take their square
roots.

The square root of 25 is 5, so we can take out the 25 as a 5 in front of
the radical.

{{{red(5)sqrt(10red(x^4)*x*red(y^2)*y)}}}

The square root of {{{x^4}}} is {{{x^2}}}, so we can take out the {{{x^4}}} as
an {{{x^2}}} in front of the radical.

{{{red(5x^2)sqrt(10x*red(y^2)*y)}}}

Finally the square root of {{{y^2}}} is {{{y}}}, so we can take out the {{{y^2}}} as just 
{{{y}}} in front of the radical.

{{{red(5x^2y)sqrt(10xy)}}}.

Edwin</pre>