How do you factor #n^2+16n+64#?

1 Answer
Mar 16, 2018

Answer:

The factored form of the polynomial is #(n+8)^2#.

Explanation:

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!