Factorise -a-1- - -a-1- -2 - ab -b-a-1- -a - b-a-1- -2 - a -a-1- -1- a - b-a-1- -2 - a - b-

The correct option is D (a 1) (2 a + b) Factorise (a 1) (a 1) ^2 + ab b (a 1) (a 1) ^2 + ab b = (a 1 (a 1)(a 1) + b(a 1) = (a 1) (1 (a 1) + b) ( ta ...

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Standard VIII

Mathematics

Factorise (a-...

Question

Factorise $(a - 1) - (a - 1)^{2} + a b - b$ .

True

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False

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$(a - 1) (a + b)$

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$(a - 1) (1 - a + b)$

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Solution

The correct option is B
False

Factorise $(a - 1) - (a - 1)^{2} + a b - b$

$(a - 1) - (a - 1)^{2} + a b - b = (a - 1 - (a - 1) (a - 1) + b (a - 1)$
= $(a - 1) (1 - (a - 1) + b)$ ( taking a-1 common)
= $(a - 1) (1 - a + 1 + b)$
= $(a - 1) (2 - a + b)$

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(ii)

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Standard VIII Mathematics

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Factorise (a−1)−(a−1)2+ab−b.

The correct option is B False Factorise (a−1)−(a−1)2+ab−b (a−1)−(a−1)2+ab−b=(a−1−(a−1)(a−1)+b(a−1) = (a−1)(1−(a−1)+b) ( taking a-1 common) = (a−1)(1−a+1+b) = (a−1)(2−a+b)

Factorise $(a - 1) - (a - 1)^{2} + a b - b$ .

The correct option is B
False

Factorise $(a - 1) - (a - 1)^{2} + a b - b$

$(a - 1) - (a - 1)^{2} + a b - b = (a - 1 - (a - 1) (a - 1) + b (a - 1)$
= $(a - 1) (1 - (a - 1) + b)$ ( taking a-1 common)
= $(a - 1) (1 - a + 1 + b)$
= $(a - 1) (2 - a + b)$