Question

4a²+b²+4ab+8a+4b+4=



a) (2a +b +2)²


b) (2a -b +2)²

c) (a +2b +2)²

d) none of them

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Answer 1

We know the algebraic identity➪

☞︎︎︎x²+y²+z²+2xy+2yz+2zx = (x+y+z)². */

Now ,

4a²+b²+4ab+8a+4b+4

= 4a²+b²+4+4ab+4b+8a

=(2a)²+b²+2²+2*2a*b+2*b*2+2*2*2a

= (2a+b+2)²

Answer 2

Answer:

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

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