How do you factor completely #12y^2+9y+8y+6#?

1 Answer
Nov 19, 2015

Answer:

Factor by grouping to find:

#12y^2+9y+8y+6 = (3y+2)(4y+3)#

Explanation:

#12y^2+9y+8y+6#

#=(12y^2+9y)+(8y+6)#

#=3y(4y+3)+2(4y+3)#

#= (3y+2)(4y+3)#

Note that the hard work / fun has already been done for you in the split #9y+8y#. Normally you would be given a quadratic like #12y^2+17y+6# to factor and have to find the split yourself.

If you are given something like #12y^2+17y+6# to factor, then one way is to multiply #12 xx 6 = 72# then try to find a pair of factors that multiply to give #72# and sum to give #17#. Once you find the pair #9#, #8# works then you can split #17y=9y+8y# and factor by grouping as we have done here.