Question:
Factorise:
(2a + 3b)2 − 16c2
Solution:
We have:
$(2 a+3 b)^{2}-16 c^{2}=(2 a+3 b)^{2}-(4 c)^{2}$
$=\{(2 a+3 b)+4 c\}\{(2 a+3 b)-4 c\}$
$=(2 a+3 b+4 c)(2 a+3 b-4 c)$
$\therefore(2 a+3 b)^{2}-16 c^{2}=(2 a+3 b+4 c)(2 a+3 b-4 c)$
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