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Are you learning how to do these? You keep posting the same type of problem.
Not at all.
Do you want to learn?
I would love to.
Take the first two terms: 3ab +3ac What do they have in common?
Do you have Skype so we don't spam this feed?
No I don't
Ok well they share "a"
What else do they share?
Good. so factor out 3a and tell me what you get
Do you understand that?
Now take the last two terms which are 2b^+2bc What do they have in common?
Factor it out
Factor out 2b. Not just 2
You said they have 2b in common and they do. So factor out 2b
I think that was my fault because I didn't get the exponent on the 2b^2
So you see that now we have 2 terms instead of 4?
And what do those two terms have in common?
Do you see that they have (b+c) in common?
Factor it out.
Factor out 9\b+c?)\
Wow, backspace is retartded
What did you say they have in common/
Then factor out b+c
They're alreadty factored out. Is your backspace working? Mine won't work on this website but others it will.
b+c times something equals the first term which is 3a(b+c)
Yes my backspace works.
Would you agree that (b+c)(3a )=3a(b+c)?
Now consider the second term 2b(b+c) (b+c) times something = 2b(b+c) What is it?
2(b+c) does not equal 2b(b+c)
Yes. So now put it all together. Look at the drawing.
And think of (b+c) as a unit.
Messed up on the drawing.
Do you see that (b+c)(3a) gives the first term and (b+c)(2b) gives the second term/
And now it is factored because it is only one term
now if you have another problem like that, you should try it yourself. And if you don't have another one, you should go back and try to understand the others that you posted. It's called "factoring by grouping"