Question

Factorise by Splitting method
25a²-4b²+28bc-49c²
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Answer 1

ANSWER

25a

2

−(4b

2

−28bc+49c

2

) splitting middle term,

=25a

2

−[4b

2

−14bc−14bc+49c

2

]

=25a

2

−[2b(2b−7c)−7c(2b−7c)]

=25a

2

−(2b−7c)

2

=(5a)

2

−(2b−7c)

2

=(5a+2b−7c)(5a−2b+7c).

Answer 2

$\huge \star \mathfrak \red{Answer!!}$

(5a²)-(2b)²-28bc-(7c)²

(5a²)-{(2b)²+28bc+(7c)²}

(5a²)-{(2b)²+2×2b×7c+(7c)²}

(5a²)-(2b+7c)² [(x+y)²=x²+2xy+y²]

{(5a²)-(2b+7c)²} {(5a²)+(2b+7c)²}

[x²-y²=(x-y) (x+y)

= (5a-2b-7c) (5a+2b+7c) 

hence proved

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