[tex]\displaystyle \frac{2x^2+4x+2}{3x^2-6x-9}=\frac{2(x^2+2x+1)}{3(x^2-2x-3)}=\frac{2(x+1)^2}{3(x^2+x-3x-3)}=\frac{2(x+1)^2}{3(x(x+1)-3(x+1))}=\\\frac{2(x+1)^2}{3(x+1)(x-3)}=\frac{2(x+1)}{3(x-3)}[/tex]