Question

Factorise:
$16(2p-3q{)}^2-4(2p-3q)$

• A. $4(2p-3q)(8p-q-1)$
• B. $(2p-3q)(8p-12q-1)$
• C. $(2p-3q)(8p-q-1)$
• D. $4(2p-3q)(8p-12q-1)$

Answer 1

The correct option is D $4(2p-3q)(8p-12q-1)$

$16(2p-3q{)}^2-4(2p-3q)$
$=4(2p-3q)\left[4\right(2p-3q)-1]$ take 4(2p-3q) common,
$=4(2p-3q)(8p-12q-1)$