
for [tex]1x^2+bx+c[/tex] to be a perfect square trinomial, [tex]c=(\frac{b}{2})^2[/tex]
or [tex]x^2+mx+m[/tex]
thereore, [tex]m=(\frac{m}{2})^2[/tex]
[tex]m=\frac{m^2}{4}[/tex]
4m=m²
m²-4m=0
(m)(m-4)=0
m=0 and m-4=0
m=0 and m=4
so you could have
x²+0x+0=(x-0)²
or
x²+4x+4=(x+2)²
answer is 3rd optio, (x+2)²