factor the trinomial..  16y²-6y-27 

Start out with this:

( y    )( y    )

There aren't too many pairs of positive
integers which have product 27, so let's
try 9 and 3 for the LASTS

( y   9)( y   3)

Now we have to think of a pair of factor
pairs that have product 16.  Try 4 and 4
for the FIRSTS:

(4y   9)(4y   3)

No that won't do, because the OUTER
product would be 12y and the INNER
product would be 36y and they don't
differ by the middle term 6y.

So let's try 2 and 8 for the FIRSTS:

(2y   9)(8y   3)

No that won't do either, because the
OUTER product would be 6y and the INNER
product would be 72y and they don't
differ by the middle term 6y.

So let's reverse the 2 and the 8:

(8y   9)(2y   3)

Now the OUTER product is 24y and the
INNER product is 18y and 24y and 
whoopee! they DO differ by the middle
term 6y.  Since the middle term is
-6x we must make the 24y negative
and the 18y positive, so we have:

(8y + 9)(2y - 3)

for the correct factorization.

Edwin