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Question

Factorise :

(i) 16a224ab     (ii) 15ab220a2b     (iii) 12x2y321x3y2


Solution

(i) 16a224ab=8a(2a3b)     {HCF of 16 and 24=8}(ii) 15ab220a2b=5ab(3b4a)     {HCF of 15 and 20=5}(iii) 12x2y321x3y2=3x2y2(4y7x)    Ans.     {HCF of 12 and 21=3}


Mathematics
Secondary School Mathematics VIII
Standard VIII

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