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

1 Answer
Jun 22, 2015

Answer:

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)#

Explanation:

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)#