Question

F(y) = 2y^3-y^2+2y-1 / y^3-y^2+y-1 and g(y) = 2y^2-3y+1/4y^2-4y+1

f(y) x g(y) =

Answer 1

`(2y^3-y^2+2y-1)/(y^3-y^2+y-1)\cdot (2y^2-3y+1)/(4y^2-4y+1)\\\\=((y^2+1)(2y-1))/((y^2+1)(y-1))\cdot ((2y-1)(y-1))/((2y-1)^2)\\\\=((y^2+1)(2y-1)^2(y-1))/((y^2+1)(2y-1)^2(y-1))=1`