We need to find two numbers that multiply to #64# (the #c# value of the quadratic) and add up to #16# (the #b# value of the quadratic).
It helps to list out the factor pairs of #64# and their sums:
#color(white){color(black)(
(1,+qquad 64 quad=65),
(2,+qquad32 quad=34),
(4,+qquad16 quad=20),
(8,+qquad8 qquad=16color(red)larr),
(16,+qquad4 qquad=20),
(32,+qquad2 qquad=34),
(64,+qquad1 qquad=65):}#
The only factors of #64# that add up to #16# are #8# and #8# (highlighted with a red arrow).
Now, split up the #n# terms in the quadratic to #8# and #8#:
#color(white)=n^2+16n+64#
#=n^2+8n+8n+64#
Now, factor the first two terms and the last two terms separately, then combine them:
#color(white)=n^2quad+quad8nquad+quad8nquad+quad64#
#=overbrace(color(red)n*n)+overbrace(color(red)n*8)+overbrace(color(blue)8*n)+overbrace(color(blue)8*8)#
#=color(red)n(n+8)+color(blue)8(n+8)#
#=(color(red)n+color(blue)8)(n+8)#
#=(n+8)^2#
That's the most factored it can get. Hope this helped!
