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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Updated on: 10-Oct-2022

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