Question 1065135
{{{(2x+r)(2x+s)=4x^2+4x+1}}}
{{{4x^2+2rx+2sx+rs}}}
{{{4x^2+2(r+s)x+rs}}}


One way to get {{{rs=1}}} is for {{{r=s=1}}}.



{{{highlight((2x+1)(2x+1))}}}
{{{4x^2+2x+2x+1}}}
{{{4x^2+4x+1}}}