Question 27735
a^2-6a-27 (this is a quadratic expression in a)
= a^2+(-9a+3a)-27 
[Observe that the product of the coeff of  a^2 and the constant term is 
(1)X(-27) = -27 = -(1x3x3x3)
and therefore we group the numerical factors 1,3,3,3 without leaving any factor  into two sets in such a way that as the  product is negative and the mid term also negative, the two sets should be  of oppostite sign with the larger numerical set having the sign of the mid term.That is our  sets 
are -(1x3x3) and (3)giving   -6 = (-9+3) so that -6a = -9a+3a and 
(-9a)X(3a) = -27a^2][The golden rule  is: split the middle term into two parts whose product is the product of the first and the last term]
=a^2-9a+3a-27
= (a^2-9a)+(3a-27)
=a(a-9) + 3(a-9)
=ap +3p where p = (a-9)
=p(a+3)
=(a-9)(a+3)
Note: Please refer to my examples, explanations and illustrations 
given under the answer for the problem 27662