Question

Can someone please help me in this factorization problem, please show your work since I want to know how to do these type of questions:

3ac - ad + 2bd - 6bc

Thanks!

Answer 1

Its actually -6bc, but how do you solve it when the letter after the common letters are different?

Answer 2

\[3ac - ad + 2bd - 6bc\]Thought of a better way. Why don't we rearrange it so that all are letters are the same between each group? \[(3ac-6bc) +(2bd-ad)\]

Answer 3

Now we pull out our common factors between each group.

Group 1 : \(3c\)

Group 2: \(d\)

Answer 4

Okay, then...

Answer 5

\[(3ac-6bc) +(2bd-ad)\]\[3c\color{red}{(a-2b)}-d\color{red}{(a-2b)}\]

Answer 6

See how we have a common factor \((a-2b)\) now?

Answer 7

Ohhh, it makes total sense, thanks for helping...

Answer 8

Are you sure?

Answer 9

Yes yes, I mean like it will be (3c-d)(a-2b) now, right?

Answer 10

Exactly :)

Answer 11

Thanks, appreciate it!

Answer 12

At first I was going to pull out a \(+d\), but if I did that our common factor of \((a-2b)\) would have been \((2b-a)\)

Answer 13

Ya, well, you have solved my problem.. Now I can solve some others....

Answer 14

Thanks once again for the explanation.

Answer 15

No problem. Good luck :)

Answer 16

@Jhannybean, Do you mind helping me in one last question?

Answer 17

\[\frac{ (x ^{2})^{2} - (y ^{2})^{2} }{(x-y)(x+y)}\]