How do you factor #12c^2+23c-9#?

1 Answer
Sep 23, 2016

Answer:

#12c^2+23c-9 = (3c-1)(4c+9)#

Explanation:

#12c^2+23c-9#

Use an AC method:

Find a pair of factors of #AC = 12*9 = 108 = 2^2*3^3# which differ by #23#

Since #23# is not divisible by #3# or #2#, one of the factors is divisible by #3^3 = 27# and one factor (possibly the same one) is divisible by #2^2 = 4#. Note that #27-4 = 23#, so we have found the required pair in #27, 4#

Use this pair to split the middle term and factor by grouping:

#12c^2+23c-9 = (12c^2+27c)-(4c+9)#

#color(white)(12c^2+23c-9) = 3c(4c+9)-1(4c+9)#

#color(white)(12c^2+23c-9) = (3c-1)(4c+9)#