Question

●hey mates

●please do this
1st and 2nd part

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#AARCHI

Answer 1

First question

a(a - 2) - b(b - 2)
Opening it

= a² - 2a -b² - 2b
=> a² - b² - 2a - 2b

Factorising a² - b² and taking 2 as common we get
(a +b)(a-b) -2(a + b)
Again taking (a + b) as common

We get, (a + b)(a -b -2)


Second question

4a² - b² + 2a +b

=> (2a)² - b² + 2a + b
Again using a² - b² = (a+b)(a-b)

=> (2a - b)(2a + b) + 1(2a + b)
Taking (2a + b) common we get
(2a + b)(2a - b + 1)


Third Question

a(a - 1) - b(b - 1)

=> a² - a - b² - b
=> a² - b² - a - b
=> (a +b)(a - b) - (a - b)

=> (a - b)(a + b -1)

Hope it helps dear friend ☺️✌️

Answer 2

Answer :

(i)

Now, a (a - 2) - b (b - 2)

= a² - 2a - b² + 2b

= a² - b² - 2a + 2b

= (a + b) (a - b) - 2 (a - b)

= (a - b) (a + b - 2),

which is the required factorization.

(ii)

Now, a (a - 1) - b (b - 1)

= a² - a - b² + b

= a² - b² - a + b

= (a + b) (a - b) - 1 (a - b)

= (a - b) (a + b - 1),

which is the required factorization.

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