Factorise 25a2−4b2+28bc−49c2
We have
25a2−4b2+28bc−49c2
=25a2−(4b2−28bc+49c2)
=(5a)2−[(2b)2−2×2b×7c+(7c)2]
=(5a)2−[(2b−7c)2] [Using identity, (a−b)2=a2−2ab+b2]
=[5a+(2b−7c)][5a−(2b−7c)] [Using identity, a2−b2=(a−b)(a+b)]
=(5a+2b−7c)(5a−2b+7c)
Factorise
(i) 4p2 − 9q2
(ii) 63a2 − 112b2
(iii) 49x2 − 36
(iv) 16x5 − 144x3
(v) (l + m)2 − (l − m)2
(vi) 9x2y2 − 16
(vii) (x2 − 2xy + y2) − z2
(viii) 25a2 − 4b2 + 28bc − 49c2