Question:
Factorise:
12(2x − 3y)2 − 16(3y − 2x)
Solution:
We have:
$12(2 x-3 y)^{2}-16(3 y-2 x)=12(2 x-3 y)^{2}+16(2 x-3 y)$
$=(2 x-3 y)\{12(2 x-3 y)+16\}$
$=(2 x-3 y)(24 x-36 y+16)$
$\therefore 12(2 x-3 y)^{2}-16(3 y-2 x)=(2 x-3 y)(24 x-36 y+16)$
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