How do you factor #2ax+6xc+ba+3bc#?

1 Answer
Jun 7, 2017

Answer:

#(a+3c)(2x+b)#

Explanation:

Let's try to find a way to group these terms.

Hmm... two of the terms have #a# and two of the terms have #c#.

Let's group the two #a# terms together, and the two #c# terms together, and then factor out #a# and #c# respectively (by the way, this is NOT the only valid way to factor this; you could actually group the terms with #x# together and the terms with #b# together and do the same thing).

#2ax + 6xc + ba + 3bc#

#2ax + ba + 6xc + 3bc#

#a(2x + b) + c(6x+3b)#

We can also pull a factor of #3# out of #6x+3b#:

#a(2x+b)+3c(2x+b)#

Now, we can use the converse of the distributive property to group #a# and #3c# together.

#(a+3c)(2x+b)#

Final Answer