Respuesta :

Tucon

 

[tex]\displaystyle\\\frac{y^3 - 4y^2 + 7y - 6}{y-2}=~~~~~\boxed{(-4y^2=-2y^2-2y^2)~~and~~(+7y=+4y+3y)}\\\\=\frac{y^3 - 2y^2-2y^2 + 4y+3y - 6}{y-2}=\\\\=\frac{y^2(y - 2)-2y(y - 2)+3(y - 2)}{y-2}=\\\\=\frac{(y - 2)(y^2-2y+3)}{y-2}=y^2-2y+3[/tex]