Factorise 4 x^{2}+y^{2}+z^{2}-4 x y-2 y z+4 x | vedclass

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Question

(A) The given expression is $4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz$.This can be written in the form of the algebraic identity $(a+b+c)^2 = a^2 + b^2 + c^

Vedclass · Accepted Answer

$(2x-y+z)^2$

Vedclass · Answer

(A) The given expression is $4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz$.This can be written in the form of the algebraic identity $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.We rewrite the expression as:$(2x)^2 + (-y)^2 + (z)^2 + 2(2x)(-y) + 2(-y)(z) + 2(2x)(z)$.Comparing this with the identity,we have $a = 2x$,$b = -y$,and $c = z$.Therefore,the expression becomes $(2x - y + z)^2$.Thus,the factors are $(2x - y + z)(2x - y + z)$.

Factorise $4 x^{2}+y^{2}+z^{2}-4 x y-2 y z+4 x z$.

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