Question #47c74

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Answer 1 · 2018-01-25T08:34:35

#3(x+1)(x-1)#

Factor out a 3 from each term in the equation:

#=3(x^2)-3(1)#

"Pull" (reverse-distribute) the 3:

#=3(x^2-1)#

#x^2-1# = #(x+1)(x-1)# because it's a difference of squares:

#=3(x+1)(x-1)#

Here's how differences of squares work:

#(x+1)(x-1)#
#=(x(x-1))+1(x-1)#
#=x^2-1x+1x-1#
#=x^2-1#

This is true for any variable x and a #(x+a)(x-a)#
#(x+a)(x-a)#
#=(x(x-a))+a(x-a)#
#=x^2-ax+ax-a^2#
#=x^2-a^2#