Question

4a²+b²+4ab+8a+4b usin identity

Answer 1

Group the terms as follows: 
4a² + (4ab + 8a) + (b² + 4b + 4) 

Factor each part separately: 
(2a)² + 4a(b + 2) + (b + 2)² 

Let u = 2a 
Let v = b + 2 

So you have: 
u² + 2uv + v² 

That's in the form of a perfect square: 
(u + v)² 

Answer: 
(2a + b+2)²

Answer 2

4a2 +b2 + 4ab +8a +4b        [a2 +2ab+b2=(a+b)2]
=(2a+b)2 + 4(2a+b)
=(2a=b)(2a+b+4)