Lakshman Patel RJIT
asked
in Quantitative Aptitude
Apr 1, 2020
retagged
Nov 12, 2020
by soujanyareddy13

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Answer is C.

Finding the factors of 1^{st} term – $4x^{^{3}}+3x^{2}y-9xy^{2}+2y^{3}$

$4x^{^{3}}-4x^{2}y+7x^{2}y-7xy^{2}-2xy^{2}+2y^{3}$

$4x^{2}(x-y)+7xy(x-y)-2y^{2}(x-y)$

$(4x^{2}+7xy-2y^{2})(x-y)$

$(4x^{2}+8xy-xy-2y^{2})(x-y)$

$4x(x+2y)-y(x+2y)$

Factors are **$(4x-y)(x+2y)$**

Finding the factors of 2nd term – $x^{2}+xy-2y^{2}$

$x^{2}-xy+2xy-2y^{2}$

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

Factors are **$(x+2y)(x-y)$**

So, HCF is the common factors between 1^{st} term and 2^{nd} term

**$(x+2y)(x-y)$**