Question

factorise the following
2x^2 +4y^2+3z^2+4√2xy-4√3yz-2√6xz​

Answer 1

Step-by-step explanation:

2x^2 +4y^2+3z^2+4√2xy-4√3yz-2√6xz

=(√2x)^2+(2y)^2+(-√3z)^2+2.√2x.2y+2.2y.-√3z+2.√2x.-√3z

=(√2x+2y-√3z)^2

Answer 2

Answer:

The factorization for given expression is $2x^{2} +4y^{2}+3z^{2} +4\sqrt{2} xy-4\sqrt{3} yz-2\sqrt{6} xz=(-\sqrt{2} x-2y+\sqrt{3} z)^{2}$

Step-by-step explanation:

The given expression in three variable which needs to be factorized is

$2x^{2} +4y^{2}+3z^{2} +4\sqrt{2} xy-4\sqrt{3} yz-2\sqrt{6} xz$

we shall factorize it in the following way as shown below

$2x^{2} +4y^{2}+3z^{2} +4\sqrt{2} xy-4\sqrt{3} yz-2\sqrt{6} xz\\=(-\sqrt{2} x)^{2} +(-2y)^{2}+(\sqrt{3} z)^{2}+2(\sqrt{2} x)(2y)+2(-2y)(\sqrt{3} z)+2(\sqrt{3} z)(-\sqrt{2} x)$

The expression obtained above is of the form $a^{2} +b^{2} +c^{2} +2ab+2bc+2ca$

We have an identity for the above expression which is mentioned below

$(a+b+c)^{2}=a^{2} +b^{2} +c^{2} +2ab+2bc+2ca$

Hence,based on this identity the above expression is

$(-\sqrt{2} x-2y+\sqrt{3} z)^{2}$

Therefore,the factorization of the given is

$2x^{2} +4y^{2}+3z^{2} +4\sqrt{2} xy-4\sqrt{3} yz-2\sqrt{6} xz=(-\sqrt{2} x-2y+\sqrt{3} z)^{2}$

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