Question 1046063
A square is already complete there.
However, {{{2x+2=2(x+1)}}} , and we love factoring.
So, you could express {{{(2x+2)^2+3}}} another way,
{{{(2x+2)^2+3=(2(x+1))^2+3=2^2(x+1)^2+3=highlight(4(x+1)^2+3)}}} ,
and that is probably what was expected.
 
If they showed you the polynomial
{{{4x^2+8x+7}}} ,
looking at the part with {{{x}}} ,
you would see a square on one {{{x}}} and another loose {{{x}}} ."Completing the square" is getting all the x's in a neat little square-of-a-binomial package.
{{{4x^2+8x+7=4x^2+8x+4+3=4(x^2+2x+1)+3=4(x+1)^2+3}}}
or
{{{4x^2+8x+7=4(x^2+2x)+7=4(x^2+2x+1-1)+7=4(x^2+2x+1)-4+7=4(x+1)^2+3}}}