Question 118450
Since we're trying to make {{{n^2+14n+c}}} a perfect square, this means {{{(a+b)^2=n^2+14n+c}}}



{{{a^2+2ab+b^2=n^2+14n+c}}} Foil {{{(a+b)^2}}}


Since we know that the first term on the right is {{{n^2}}}, this means {{{a^2=n^2}}} which tells us that {{{a=n}}}



{{{n^2+2nb+b^2=n^2+14n+c}}} Plug in {{{a=n}}}



So this means that {{{2nb=14n}}} (notice how both of these terms only have 1 "n") and {{{b^2=c}}} (notice how both of these terms have no "n" terms)



So let's use {{{2nb=14n}}} to solve for b



{{{2nb=14n}}} Start with the given equation



{{{cross(2n/2n)b=14n/2n}}} Divide both sides by 2n to isolate b




{{{b=7}}} Divide 






{{{n^2+14n+7^2}}} Now plug in {{{b=7}}} into the original polynomial {{{n^2+14n+c}}}




{{{n^2+14n+49}}} Now square 7 to get 49



So this means that the perfect square is {{{n^2+14n+49}}} 




Check:



Notice how if we factor {{{n^2+14n+49}}} we get {{{(n+7)^2}}} which shows us that {{{n^2+14n+49}}} is a perfect square. So our answer is verified.