How do you factor 2ax+6xc+ba+3bc2ax+6xc+ba+3bc?

1 Answer
Jun 7, 2017

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

Explanation:

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

Hmm... two of the terms have aa and two of the terms have cc.

Let's group the two aa terms together, and the two cc terms together, and then factor out aa and cc respectively (by the way, this is NOT the only valid way to factor this; you could actually group the terms with xx together and the terms with bb together and do the same thing).

2ax + 6xc + ba + 3bc2ax+6xc+ba+3bc

2ax + ba + 6xc + 3bc2ax+ba+6xc+3bc

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

We can also pull a factor of 33 out of 6x+3b6x+3b:

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

Now, we can use the converse of the distributive property to group aa and 3c3c together.

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

Final Answer