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)
Answer:
(y+7)/(4y+1).
Step-by-step explanation:
If we factor all the polynomials we get:
(4y-1)(y-3) . (2y+3)(y+3) . (y-4)(y+7)
-------------- ---------------- -------------
(2y+3)(y-4) (4y-1)(4y+1) (y-3)(y+3)
we see that (4y-1), (y -3), (2y+3), (y+3) and (y - 4) are in the numerators and denominators so we can cancel them out.
This leaves us with:
(y + 7)
--------
(4y + 1)
