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

$\dfrac{2y^3-y^2+2y-1}{y^3-y^2+y-1}\cdot \dfrac{2y^2-3y+1}{4y^2-4y+1}\\\\=\dfrac{(y^2+1)(2y-1)}{(y^2+1)(y-1)}\cdot \dfrac{(2y-1)(y-1)}{(2y-1)^2}\\\\=\dfrac{(y^2+1)(2y-1)^2(y-1)}{(y^2+1)(2y-1)^2(y-1)}=1$