6a²+8a+2

First factor out common factor 2

2[3a²+4a+1]

You may be able to go straight from there to 
the final answer 2(a+1)(3a+1) by factoring 
the simple way.  But if not:

Multiply 3 by 1 get 3. Think of 3·1=3 and 3+1=4

Then write 4a as 3a+1a

2[3a²+3a+1a+1]

Factor the first two terms in the bracket:

2[3a(a+1)+1(a+1)]

2[(a+1)(3a+1)]

2(a+1)(3a+1)

Edwin