How do you factor #(b - 3)^2 - (a + 1)^2 #?

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Answer 1 · 2015-06-22T23:18:54

This is a difference of squares, so it factors as

#(b-3)^2 - (a+1)^2#

#= ((b-3)-(a+1))((b-3)+(a+1))#

#=(b-a-4)(b+a-2)#

For any #p# and #q#, we have:

#p^2-q^2 = (p-q)(p+q)#

Putting #p=(b-3)# and #q=(a+1)#, we have:

#(b-3)^2-(a+1)^2#

#= p^2-q^2#

#= (p-q)(p+q)#

#= ((b-3)-(a+1))((b-3)+(a+1))#

#=(b-a-4)(b+a-2)#