How do you factor #(9a+3b-c)^2-(3a-2b+c)^2 #?

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Answer 1 · 2018-04-22T04:07:43

#a^2 - b^2 = (a+b)(a-b)#

Using that will give you #(6a+5b-2c)(12a+b)#

Answer 2 · 2018-04-22T04:09:46

Using the difference of two squares, we get # (12 a + b)(6a+5b-2c)#.

In general the difference of two squares factors easily:

#x^2 - y^2 = (x+y)(x-y)#

#(9a+3b-c)^2-(3a-2b+c)^2 #

#= (9a + 3b -c + 3a -2b +c)(9a + 3b -c - 3a +2b -c) #

#= (12 a + b)(6a+5b-2c)#