factorise, 25a^2-4b^2+28bc-49c^2
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25a
2
−4b
2
−28bc−49c
2
=(5a)
2
−(2b)
2
+28bc−(7c)
2
=(5a)
2
−{(2b)
2
+2×2b×7c−(7c)
2
}
∵[(a+b)
2
=a
2
+b
2
+2ab]
=(5a)
2
−(2b+7c)
2
∵[a
2
−b
2
=(a−b)(a+b)]
={(5a)−(2b+7c)}{(5a)+(2b+7c)}
=(5a−2b−7c)(5a+2b+7c)
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Answered by
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Answer:
Answer
25a
2
−4b
2
−28bc−49c
2
=(5a)
2
−(2b)
2
+28bc−(7c)
2
=(5a)
2
−{(2b)
2
+2×2b×7c−(7c)
2
}
∵[(a+b)
2
=a
2
+b
2
+2ab]
=(5a)
2
−(2b+7c)
2
∵[a
2
−b
2
=(a−b)(a+b)]
={(5a)−(2b+7c)}{(5a)+(2b+7c)}
=(5a−2b−7c)(5a+2b+7c)
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