Factorise 4(a – 1)2 – 4(a – 1) – 3

(i) 4(a – 1)² – 4(a – 1) – 3

(ii) 1 – 2a – 2b – 3(a + b)²

Answer :

(i) 4(a – 1)² – 4(a – 1) – 3

Above terms can be written as,

4(a – 1)² – 6(a – 1) + 2(a – 1) – 3

Take out common in all terms we get,

2(a – 1) [2(a – 1) – 3] + 1[2(a – 1) – 3]

(2(a – 1) – 3) (2(a – 1) + 1)

(2a – 2 – 3) (2a – 2 + 1)

(2a – 5) (2a – 1)

(ii) 1 – 2a – 2b – 3(a + b)²

1 – 2a – 2b – 3(a + b)²

Above terms can be written as,

1 – 2(a + b) – 3(a + b)²

1 – 3(a + b) + (a + b) – 3(a + b)²

Take out common in all terms we get,

1(1 – 3(a + b)) + (a + b) (1 – (a + b))

(1 – 3(a + b)) (1 + (a + b))

(1 – 3a + 3b) (1 + a + b)

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