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Factorise $25a^2-4b^2+28bc-49c^2$.
Given:
The given expression is $25a^2-4b^2+28bc-49c^2$.
To do:
We have to factorise the given expression.
Solution:
$25a^2-4b^2+28bc-49c^2=(5a)^2-(2b)^2+28bc-(7c)^2$
$=(5a)^2-[(2b)^2-2(2b)(7c)+(7c)^2]$
$=(5a)^2-[(2b-7c)^2]$
$=(5a)^2-(2b-7c)^2$
$=(5a+2b-7c)[5a-(2b-7c)]$ (Since $x^2-y^2=(x+y)(x-y)$)
$=(5a+2b-7c)(5a-2b+7c)$
Factors of $25a^2-4b^2+28bc-49c^2$ are $(5a+2b-7c)$ and $(5a-2b+7c)$.
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