How do you factor #(n+4)^2 - y^2#?

Question

I see that this is a difference of two squares, but how do I continue factoring it after I get to #((n+4)+y)((n+4)-y)#?
Or, is this as far as it can be factored?

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Answer 1 · 2017-11-11T13:02:31

#(n+y+4)(n-y+4)#

Given #(n+4)^2 - y^2#

Let #x = n + 4#

#:. = x^2 - y^2#

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

Substituting for value of x,

#= (n + 4 +y)(n + 4 -y)#