Question #462d9

Question

1890 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-25T16:06:57

#3[(9(3x-1) +4(2y+3))(9(3x-1)-4(2y+3))]#

Always look for a common factor first. # is a common factor here.

#243(3x-1)^2-48(2y+3)^2#
#=3[81(3x-1)^2 - 16(2y+3)^2]#

Inside the brackets we have the difference of 2 squares.
This will be easier to recognise if we let (3x-1) = p
and (2y+3) = q.

#3[81(3x-1)^2 - 16(2y+3)^2]# becomes
#3[81p^2 - 16q^2]#

=#3[(9p +4q)(9p-4q)]#

Replace p and q by their real values to get:

#3[(9(3x-1) +4(2y+3))(9(3x-1)-4(2y+3))]#