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Factorise

(i) (x2 − 2xy + y2) − z2

(ii) 25a2 − 4b2 + 28bc − 49c2

1 Answer

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Best answer

(i) (x2 − 2xy + y2) − z

= (x − y)2 − (z)2 [(a − b)2 = a2 − 2ab + b2]

= (x − y − z) (x − y + z) [a2 − b2 = (a − b) (a + b)]

(ii) 25a2 − 4b2 + 28bc − 49c2 

= 25a2 − (4b2 − 28bc + 49c2

= (5a)2 − [(2b)2 − 2 × 2b × 7c + (7c)2]

= (5a)2 − [(2b − 7c)2]

[Using identity (a − b)2 = a2 − 2ab + b2

= [5a + (2b − 7c)] [5a − (2b − 7c)] 

[Using identity a2 − b2 = (a − b) (a + b)] 

= (5a + 2b − 7c) (5a − 2b + 7c)

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