How do you factor by grouping #6c^2+29c-42#?

2 Answers
Apr 22, 2015

#6c^2+29c-42#

We are looking for factors, #m# and #n#, of #6#
and
factors, #p# and #q#, of #(-42)#
such that
#mq+np=29#

  • this is because #(mc+p)*(nc+q) = mnc^2+(mq+np)c +pq#

factors of 6 #= {(1,6),(2,3)}# plus their negatives
factors of (-42) # = {(6,-7),(3,-14),(2,-21),(1,-42)} and their negatives

A bit of testing combinations comes up with the pairs
#(p,q) = (6,1)#
and
#(m,n)=(-7,6)#
that satisfy the requirement for the middle term

#6c^2+29c-42 = (6c-7)(1c+6)#

Apr 22, 2015

Multiply #6 xx -42 = -252#

Now we look for factor of #-252# that add to get #29#

Because the product is negative and the sum is positive, we want the smaller number to be negative and the larger positive:
-1 252 does not add up to 29
-2 126 does not add up to 29
-3 84
-4 63
-5 is not a factor
-6 42

-7 36 does add up to 29

In #6c^2 + 29c -42# split the #29c# into #-7c+36c#
(or #36c-7c# -- either will work)

#6c^2 + 29c -42 = 6c^2 + -7c + 36c -42#

#color(white)"ssssssssssssssss"# #=(6c^2 -7c) + (36c -42)#

#color(white)"ssssssssssssssss"# #=c(6c -7) + 6(6c - 7)#

#color(white)"ssssssssssssssss"# #=(c+ 6)(6c - 7)#