Question:
Factorise:
25a2 − 4b2 + 28bc − 49c2
Solution:
We have:
$25 a^{2}-4 b^{2}+28 b c-49 c^{2}=25 a^{2}-\left(4 b^{2}-28 b c+49 c^{2}\right)$
$=(5 a)^{2}-(2 b-7 c)^{2}$
$=\{5 a+(2 b-7 c)\}\{5 a-(2 b-7 c)\}$
$=(5 a+2 b-7 c)(5 a-2 b+7 c)$
$\therefore 25 a^{2}-4 b^{2}+28 b c-49 c^{2}=(5 a+2 b-7 c)(5 a-2 b+7 c)$
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