(i) (a2 – b2) (c2 – d2) – 4abcd
(ii) 4x2 – y2 – 3xy + 2x – 2y
Answer “
(i) (a2 – b2) (c2 – d2) – 4abcd
a2(c2 – d2) – b2 (c2 – d2) – 4abcd
a2c2 – a2d2 – b2c2 + b2d2 – 4abcd
a2c2 + b2d2 – a2d2 – b2c2 – 2abcd – 2abcd
Rearranging the above terms, we get,
a2c2 + b2d2 – 2abcd – a2d2 – b2c2 – 2abcd
We know that, (a + b)2 = a2 + 2ab + b2 and (a – b)2 = a2 – 2ab + b2
(ac – bd)2 – (ad – bc)2
(ac – bd + ad – bc) (ac – bd – ad + bc)
(ii) 4x2 – y2 – 3xy + 2x – 2y
4x2 – y2 – 3xy + 2x – 2y
Above terms can be written as,
x2 + 3x2 – y2 – 3xy + 2x – 2y
Rearranging the above terms, we get,
(x2 – y2) + (3x2 – 3xy) + (2x – 2y)
We know that, a2 – b2 = (a + b) (a – b) and take out common terms,
(x + y) (x – y) + 3x(x – y) + 2(x – y)
(x – y) [(x + y) + 3x + 2]
(x – y) (x + y + 3x + 2)
(x – y) (4x + y + 2)
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