Answer
$\frac{4}{y+5}$ (and $y\ne5$).
Work Step by Step
The product of a binomial sum and a binomial difference is: $(A+B)(A-B)=A^2-B^2$.
Hence here: $\frac{4y-20}{y^2-25}=\frac{4(y-5)}{y^2-5^2}=\frac{4(y-5)}{(y+5)(y-5)}=\frac{4}{y+5}$ (and $y\ne5$).
