Factor the expression. Factor out all common factors, if they exist.?

= - 6abc - 6ac² + 3bc + 3c²
= 3c(- 2ab - 2ac + b + c)
= 3c(- 2a + 1)(b + c)
= 3c(2a - 1)(b + c)

Answer: 3c(2a - 1)(b + c) or 3c(- 2a + 1)(b + c)

3c [ - 2ab - 2ac + b + c ]
3c [ (- 2a)(b + c) + (b + c) ]
3c [ (b + c) (1 - 2a) ]

Why are you spamming the question section with such easy questions. Ask one or two and use the answers to help you understand how to do the question on your own.

Mass spamming what is 1+1, 2+3, ... . Is just stupid and very ignorant. Respect other people using this service.

factor by group


3c^2 - 6ac^2 = 3c^2 (1 - 2a)
3bc- 6abc = 3bc ( 1 - 2a)

thus you will have 3bc ( 1 - 2a) + 3c^2 (1 - 2a)

your common factor is: (1 - 2a)

therefore, you will have


(1 - 2a) (3bc + 3c^2) or 3c ( b +c )( 1 - 2a)

-6abc - 6ac² + 3bc + 3c²

(-6abc - 6ac²) + (3bc + 3c²)

-6ac(b + c) + 3c(b + c)

(-6ac + 3c)(b + c)

hope this helped!

( 3 - 6 a c )( b c + c^2 )

-6abc - 6ac^2 + 3bc + 3c^2
= 3(-2abc - 2ac^2 + bc + c^2)
= 3c(-2ab - 2ac + b + c)
= 3c(-2a + 1)(b + c)