Answer: f(y) x g(y) = 1
Step-by-step explanation:
[tex]\frac{2y^3-y^2+2y-1}{y^3-y^2+y-1}\cdot \frac{2y^2-3y+1}{4y^2-4y+1}\\\\= \frac{4y^5-8y^4+9y^3-9y^2+5y-1}{4y^5-8y^4+9y^3-9y^2+5y-1} \\\\= \frac{(y-1)(2y-1)(2y-1)(y^2+1)}{(y-1)(2y-1)(2y-1)(y^2+1)} \\\\= 1[/tex]
