Factorise: 16(2p − 3q)2 − 4(2p − 3q)

Question:

Factorise:
16(2p − 3q)2 − 4(2p − 3q)

Solution:

 We have:

$16(2 p-3 q)^{2}-4(2 p-3 q)=(2 p-3 q)\{16(2 p-3 q)-4\}$

$=(2 p-3 q)(32 p-48 q-4)$

$\therefore 16(2 p-3 q)^{2}-4(2 p-3 q)=(2 p-3 q)(32 p-48 q-4)$

 

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