(10y^2-13y-30)/(2y^5-50y^3)
= (10y^2-25y+12y-30)/(2y^3(y^2-25)
= (5y(2y-5)+6(2y-5))/(2y^3(y+5)(y-5))
= ((5y+6)(2y-5))/(2y^3(y+5)(y-5))
Hence, f(y)=(10y^2-13y-30)/(2y^5-50y^3)<0
hArrf(y)=((5y+6)(2y-5))/(2y^3(y+5)(y-5))<0
:. zeros of numerator and denominators are
{-5,-6/5,0,5/2,5} and they divide real number line in 6 intervals.
In -oo< y< -5, all monomials are negative and hence f(y)<0
In -5< y< -6/5, while (y+5) is positive, all other monomials are negative and hence f(y)>0
In -6/5< y< 0, while (y+5) and (5y+6) are positive, all other monomials are negative and hence f(y)<0
In 0< y< 5/2, while (y+5), (5y+6) and y are positive, (2y-5) and (y-5) are negative and hence f(y)>0
In 5/2< y< 5, while (y+5), (5y+6), y and (2y-5) are positive, (y-5) is negative and hence f(y)<0
and in 5< y< oo, all monomials are positive and hence f(y)>0
Hence solution is -oo< y< -5 or -6/5< y< 0 or 5/2< y< 5.
graph{(10x^2-13x-30)/(2x^5-50x^3) [-10, 10, -2, 2]}
