Question

factorise: 25a²-4b²+28bc-49c².​

Answer 1

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Answer 2

Given:

f(a,b,c)=25a²–4b²+28bc–49c²

Factorisation:

»25a²–4b²+28bc–49c²

»(5a)²–(2b)²–28bc–(7c)²

»(5a)²–{(2b)²+28bc+(7c)²}

»(5a)²–{(2b)²+(2×2b×7c)+(7c)²}

»(5a)²–(2b+7c)² [∵(a+b)²=a²+2ab+b²]

»[(5a)–(2b+7c)][(5a)+(2b+7c)]

[∵a²–b²=(a–b) (a+b)]

»(5a–2b–7c)(5a+2b+7c) which are the required factors.