Question 943590
What said here is not quite right:
<s>You should be able to find the combination of values to pick the parts of that perfect square through simple trials of possible combinations.  You see "-6y", so you expect to use two negative number constants.</s>


My mistake:  {{{cross((3y-2)^2)}}}


If you are trying to factor the original expression, try looking for combinations of binomials that will work.

(3y_______1)(3y_______6)
3y and 18y
not work


(3y______2)(3y______3)
6y and 9y
not work


(9y_______1)(y________6)
54 and 1
not work


Anything else?


(9y_______6)(y_____1)
9 and 6
not work


(9y______2)(y_______3)
27 and 2
not work


(9y________6)(y_______1)
9 and 6
not work


(9y______3)(y_______2)
18 and 3
not work


THIS should be possible, should be obvious:
{{{highlight(3(3y^2-2y+2))}}}
and you could again check possible combinations and test them, but maybe using Discriminant is easier.
{{{(-2)^2-4*3*2}}}
{{{4-24<0}}}
The discriminant is negative, so NO real roots, meaning {{{3y^2-2y+2}}} IS NOT FACTORABLE.


Finally, YOUR answer is correct.