How do you factor #12r^2 - 7r - 12# by grouping?

1 Answer
Aug 18, 2016

Answer:

#12r^2-7r-12=(4r-3)(3r-4)#

Explanation:

We need to find how to split the middle term. To do that, use an AC method:

Look for a pair of factors of #AC = 12*12 = 144# which differ by #B=7#

The pair #16, 9# works, so we find:

#12r^2-7r-12#

#=12r^2-16r+9r-12#

#=(12r^2-16r)+(9r-12)#

#=4r(3r-4)+3(3r-4)#

#=(4r-3)(3r-4)#