Question

6 Factorice (x+2y + 4z)² using suitable identity


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Answer 1

Answer:

factorise (x+2y+4z)² using suitable identity

(a+b+c)²=a²+b²+c²+2ab+2bc+2ca

=x²+(2y)²+(4z)²+2x(2y)+2(2y)(4z)+2(4z)(x)

=x²+4y²+16z²+4xy+16yz+8xz