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

Jun 7, 2017

$\left(a + 3 c\right) \left(2 x + b\right)$

#### 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).

$2 a x + 6 x c + b a + 3 b c$

$2 a x + b a + 6 x c + 3 b c$

$a \left(2 x + b\right) + c \left(6 x + 3 b\right)$

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

$a \left(2 x + b\right) + 3 c \left(2 x + b\right)$

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

$\left(a + 3 c\right) \left(2 x + b\right)$