How do you factor #243(3x - 1)^2 - 48(2y + 3)^2#?

1 Answer
Aug 18, 2016

Answer:

Use the difference of squares property to get #3(27x+8y+3)(27x-8y-21)#.

Explanation:

What should always jump out at you in a factoring question containing a minus sign and stuff squared is difference of squares:
#a^2-b^2=(a-b)(a+b)#

But the 243 and 48 kind of kill that idea, because they aren't perfect squares. However, if we factor out a #3#, we have:
#3(81(3x-1)^2-16(2y+3)^2)#

Which can be rewritten as:
#3((9(3x-1))^2-(4(2y+3))^2)#

Now we can apply difference of squares, with:
#a=9(3x-1)#
#b=4(2y+3)#

Doing so gives:
#3((9(3x-1))^2-(4(2y+3))^2)#
#=3((9(3x-1)+4(2y+3))(9(3x-1)-4(2y+3))#

Let's get rid of some parentheses by distributing:
#3((9(3x-1)+4(2y+3))(9(3x-1)-4(2y+3))#
#=3(27x-9+8y+12)(27x-9-8y-12)#

Finally, collect terms:
#3(27x-9+8y+12)(27x-9-8y-12)#
#=3(27x+8y+3)(27x-8y-21)#