Question

factorise, 25a^2-4b^2+28bc-49c^2

Answer 1

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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Answer 2

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)