For this, I will be factoring by grouping to solve for y.
Firstly, factor y^3 + 5y^2 and -9y - 45 separately, Make sure that they have the same quantity on the inside:
![y^2(y+5)-9(y+5)=0](https://img.qammunity.org/2019/formulas/mathematics/middle-school/znnp70er92ggfhm775zzogg33fw0o5c1mm.png)
Now you can rewrite it as
. However we are not finished factoring.
Now with y^2 - 9, we will apply the difference of squares rule, which is
. Apply it as such:
![(y+3)(y-3)(y+5)=0](https://img.qammunity.org/2019/formulas/mathematics/middle-school/oungc7y3yetlvhby2vspb2sbhjgucwi8pt.png)
Now that it is completely factored, we can apply the Zero Product Property and solve for y:
![y+3=0\\y=-3\\\\y-3=0\\y=3\\\\y+5=0\\y=-5](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ga2lr9q35xe158ubohmp02serv4yll7o7l.png)
In short, y = -5, -3, and 3.