Question:
Factorise:
6ab − b2 + 12ac − 2bc
Solution:
By suitably arranging the terms:
$6 a b-b^{2}+12 a c-2 b c=6 a b+12 a c-b^{2}-2 b c$
$=(6 a b+12 a c)-\left(b^{2}+2 b c\right)$
$=6 a(b+2 c)-b(b+2 c)$
$=(b+2 c)(6 a-b)$
$\therefore 6 a b-b^{2}+12 a c-2 b c=(b+2 c)(6 a-b)$
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