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Factorise :

98(a+b)2298(a + b)^2 - 2

Factorisation

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Answer

98(a+b)22=2[49(a+b)21]=2[(7(a+b))212]=2[7(a+b)1][7(a+b)+1]=2(7a+7b1)(7a+7b+1)98(a + b)^2 - 2\\[1em] = 2[49(a + b)^2 - 1]\\[1em] = 2[(7(a + b))^2 - 1^2]\\[1em] = 2[7(a + b) - 1][7(a + b) + 1]\\[1em] = 2(7a + 7b - 1)(7a + 7b + 1)

Hence, 98(a+b)22=2(7a+7b1)(7a+7b+1)98(a + b)^2 - 2 = 2(7a + 7b - 1)(7a + 7b + 1).

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