Factorise:25a2−4b2+28bc−49c2
25a2–4b2+28bc–49c2
=(5a)2–(4b2–28bc+49c2)
=(5a)2–[(2b)2–2(2b)(7c)+(7c)2] [Using x2–2xy+y2=(x–y)2 ]
=(5a)2–(2b–7c)2
=(5a–2b+7c)(5a+2b–7c) [Using x2–y2=(x–y)(x+y)]
Hence, 25a2–4b2+28bc–49c2=(5a–2b+7c)(5a+2b–7c)