\(=\dfrac{y^4\left(y-2\right)+2y^2\left(y-2\right)-3\left(y-2\right)}{y^2-2y+4y-8}\\ =\dfrac{\left(y-2\right)\left(y^4+2y^2-3\right)}{\left(y-2\right)\left(y+4\right)}=\dfrac{y^4-y^2+3y^2-3}{y+4}\\ =\dfrac{\left(y^2+3\right)\left(y^2-2\right)}{y+4}\)