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Question

Factorize: (1x2)(1y2)+4xy

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Solution

(1x2)(1y2)+4xy
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=1x2y2+x2y2+4xy
=1x2y2+x2y2+2xy+2xy
=1+x2y2+2xy(x2+y22xy)
=(1+(xy))2(yx)2[a2b2]
=(1+xy+yx)(1+xyy+x)a2b2=(a+b)(ab)

1172674_1246569_ans_699b5d59b9f344f89032adec5dd068f3.jpeg

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