2011/041) factoring by diminishing power $6bc - 9c² - 12cd - 8be + 12ce +16de$

one of the factoring that I saw in the net was factor

$6bc - 9c² - 12cd - 8be + 12ce +16de$

...the "clue" you need to recognize is that the coefficients of the first three terms have a common factor...and the coefficients of the last three terms have a common factor...and, the ratios are constant...so, group accordingly...

$(6bc - 9c^2 - 12cd) - (8be - 12ce - 16de) = 3c(2b - 3c - 4d) - 4e(2b - 3c - 4d) =$
$(2b - 3c - 4d)(3c - 4e)$

the above is correct and it was luck that grouping was there but what is luck is not there
then see that highest power of c is 2 and keep them in descending order

$6bc - 9c² - 12cd - 8be + 12ce +16de$
$= - 9c² + 6bc - 12cd + 12ce - 8be + 16de$
now you can factor as quadratic in c eliminating the luck factor
$= - 9c² - + 6c( b - 2d + 2e) -8e(b- 2d)$
$= - 9c² - + 6c(( b - 2d) + 2e)) -8e(b- 2d)$
letting b - 2d = a we get
$= - 9c^2 + 6c(a + 2e) - 8ea$
$= - 9c^2 + 6ca + 12ec - 8ea$
$= - 3c(3c - 2a) + 4e(3c-2a)$
$= (3c-2a)(4e - 3c)$
$= (3c-2b+4d)(4e-3c)$
which is same as 1st one

I do not mean to say that 2nd one is preferable to 1st but 2nd one can be used when 1st one does not work