Question

Factorise:- a²+b²+4c²+2ab-4bc-4ca ​

Answer 1

Answer:

4a

2

+b

2

+c

2

−4ab+2bc−4ca

=(2a)

2

+(−b)

2

+(−c)

2

+2(2a)(−b)+2(−b)(−c)+2(2a)(−c)

We know that x

2

+y

2

+z

2

+2xy+2yz+2xz=(x+y+z)

2

Here, x=2a, y=−b and z=−c

So, 4a

2

+b

2

+c

2

−4ab+2bc−4ca

=(2a−b−c)

2

=(2a−b−c)(2a−b−c)