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Question

Factorise :

(i) 14x3+21x4y28x2y2 (ii) 510t+20t2

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Solution

{HCF of 14,21,28=7}

(i) 14x3+21x4y28x2y2

=7x2(2x+3x2y4y2) [Taking 7x2 as common]


{HCF of 5,10,20=5}
(ii) 510t+20t2

=5(1+2t4t2) [Taking 5 as common]


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