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Question

Factorise :
(i)16a2254a2
(ii)16a2bb16a2
(iii)100(x+y)281(a+b)2
(iv)(x1)2(x2)2

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Solution

We have,
(i)16a2254a2
=(4a)2(52a)2=(4a+52a)(4a52a)
(ii)16a2bb16a2=b(16a2116a2)
=b{(4a)2(14a)2}
=b(4a+14a)(4a14a)

(iii)100(x+y)281(a+b)2={10(x+y)}2{9(a+b)}2
={10(x+y)+9(a+b)}{10(x+y)9(a+b)}
=(10x+10y+9a+9b)(10x+10y9a9b)

(iv)(x1)2(x2)2={(x1)+(x2)}{(x1)(x2)}
=(2x3)(x1x+2)
=(2x3)×1
=2x3

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