Answer:
(y^2+3y-28)/(4y^2-15y-4)
Step-by-step explanation:
(4y-1)(y-3)(2y+3)(y+3)(y+7)(y-3)/(2y+3)(y-4)(4y+1)(4y-1)(y-3)(y+3)
= (y+7)(y-3)/(y-4)(4y+1)
= (y^2+3y-28)/(4y^2-15y-4)
the 2 one is a linear and the rest are non
