Question 1085953


For what real values of {{{c}}} is {{{x^2 + 16x + c}}} the square of a binomial?

recall the square of a binomial:{{{(a+b)^2=a^2+2ab+b^2}}}

you have {{{x^2 + 16x + c}}}-> {{{a=1}}}, {{{2ab=16}}}, and {{{b^2=c}}}

find {{{b}}}-> {{{b=16/2a}}}-> {{{b=16/(2*1)}}}-> {{{b=8}}}

then, {{{8^2=c}}}->{{{c=64}}}

{{{x^2 + 16x + 64}}}

{{{x^2 + 16x + 8^2}}}

{{{(x + 8)^2}}}