How do you factor by grouping #3p^2 - 2p - 5#?

1 Answer
May 29, 2015

#3p^2-2p-5 = 3p^2+3p-5p-5#

#= (3p^2+3p)-(5p+5)#

#=3p(p+1)-5(p+1)#

#=(3p-5)(p+1)#

The trick here is how to split the middle term into two so that the ratio of the resulting 1st and 2nd term is the same as the ratio of the 3rd and 4th term. In this particular example it was easy to spot, but in general you may like to use a variant of the AC Method to find a suitable pair to use.

To see how this works, identify the coefficients of the terms ignoring the signs: #A=3#, #B=2#, #C=5#. Noticing that the sign of the last, constant, term is #-#, we then look for pairs of factors of #AC = 15# whose difference is #B#. The pair #5, 3# satisfies:
#5 xx 3 = 15 = AC# and #5 - 3 = 2 = B#. So the pair of coefficients to split the middle term into is #5# and #3#, as we have done above.