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6aČ+8a+2
First factor out common factor 2
2[3aČ+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[3aČ+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