Question 75090
You could try this:
{{{9x^2+12xy+4y^2-25}}} Try grouping the terms as follows:
{{{(9x^2+12xy+4y^2)-(25))}}} Do you notice that, in the first group, the first and last terms are perfect squares>  This suggests that this trinomial might be a perfect square. Let's try it.
{{{9x^2+12xy+4y^2 = (3x+2y)(3x+2y)}}} = {{{(3x+2y)^2}}} Yes...a perfect square! So we can rewrite the original expression as:
{{{(9x^2+12xy+4y^2)-(25) = (3x+2y)^2-(5)^2}}} Now you can see that we have the difference of two squares which can be factored thusly: {{{A^2-B^2 = (A+B)(A-B)}}} Applying this to your expression, we get:
{{{(3x+2y)^2-(5)^2 = ((3x+2y)+5)((3x+2y)-5)}}}
The final answer looks like:
{{{9x^2+12xy+4y^2-25 = (3x+2y+5)(3x+2y-5)}}}
Check: Using the FOIL method, we'll multiply the factors to see if we get the original expression back.
{{{(3x+2y+5)(3x+2y-5) = 9x^2+6xy-15x+6xy+4y^2-10y+15x+10y-25}}} Simplifying this, we get:{{{9x^2+12xy+4y^2-25}}} the original expression.