Question 621893
{{{ 4x^2}}} is the square of {{{2x}}}
{{{(2x+b)^2=4x^2+4bx+b^2}}}
If {{{4bx=12x}}} 
no matter what value x takes, it must be
{{{4b=12}}} <---> {{{b=3}}} and then
{{{(2x+b)^2=(2x+3)^2=4x^2+12x+3^2=4x^2+12x+9}}}
So {{{highlight(c=9)}}}