5 (3v + 5p) (3v + 5p)

5 (3v + 5p)²

5(9v^2 + 30vp + 25p^2)

Inside, notice that 9v^2 = (3v)^2 and 25p^2 = (5p)^2. This leads me to suspect that the polynomial is a perfect square. Let’s check that out:

(3v + 5p)^2 = 9v^2 + 15vp + 15vp + 25p^2 (using FOIL)

= 9v^2 + 30vp + 25p^2

and we have it. So it factors to

5(3v + 5p)^2

We know we’re done because (a) there are no more common factors and (b) all of our factors are linear – i.e. the largest power on any variable is 1. You won’t always go down that far, but when you do, you know there’s nowhere else to go.

The letters can’t go out because they don’t have any in common between the three, but they all have the common 5.