Factorise:

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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