1
Question
25a2 − 4b2 + 28bc − 49c2
25a2 − 4b2 + 28bc − 49c2
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Solution
Dear Student!
25a2 – 4b2– 28bc – 49c2
= (5a)2 – (2b)2– 28bc – (7c)2
= (5a)2 – {(2b)2 + 28bc + (7c)2}
= (5a)2 – {(2b)2 + 2 × 2b × 7c + (7c)2}
= (5a)2 – (2b + 7c)2 [(x + y)2 = x2 + 2xy + y2]
= {(5a) – (2b + 7c)} {(5a) + (2b + 7c)} [x2 – y2 = (x – y) (x + y)]
= (5a – 2b – 7c) (5a + 2b + 7c)
Cheers!
Dear Student!
25a2 – 4b2– 28bc – 49c2
= (5a)2 – (2b)2– 28bc – (7c)2
= (5a)2 – {(2b)2 + 28bc + (7c)2}
= (5a)2 – {(2b)2 + 2 × 2b × 7c + (7c)2}
= (5a)2 – (2b + 7c)2 [(x + y)2 = x2 + 2xy + y2]
= {(5a) – (2b + 7c)} {(5a) + (2b + 7c)} [x2 – y2 = (x – y) (x + y)]
= (5a – 2b – 7c) (5a + 2b + 7c)
Cheers!
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