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anonymous
 5 years ago
I'm trying to figure out the steps to factoring, such as the 1st problem...
2x^2y6xy^2
anonymous
 5 years ago
I'm trying to figure out the steps to factoring, such as the 1st problem... 2x^2y6xy^2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Each of the terms \(2x^2y\) and \(6xy^2\) are products of some factors. The first step is to see what factors they have in common.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What are the factors you can see in both those terms?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02 is one of the factors they have in common. There are 2 more though.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I was thinking 2 because 2 goes into 2 one time, and 2 goes into 6, 3 times. The only other common factor I can see if it has something to do with x & y.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So would I divide both sides by 2xy?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, and bring that 2xy out in front.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, sorry. Factor out 2xy from \(2x^2y\) and \(6xy^2\) You should have 2xy(x  3y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And you can see if you redistribute the 2xy that you will get what you started with.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, it's making a little bit more sense. How does the 2nd part of the answer become (x3y) where the x is separated from the y?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Each of those terms x and 3y are what's left over when you pull out the 2xy factor from both of them.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just like factoring (15  25) into 5(35)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, I think I got it. You've been awesome. :) Thank you for your help.
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