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anonymous
 3 years ago
How do you factor 5ab2b squared2b+5a?
anonymous
 3 years ago
How do you factor 5ab2b squared2b+5a?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0try rearranging them \[\implies 5ab + 5a  2b^2  2b\] do you see how to factor it now? or no?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No... Like I want to be taught the process to fully factor it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well the first step is to rearrange it (like i did)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the next step...is to group the terms like this \[\implies (5ab + 5a) + (2b^2  2b)\] still don't see the factor?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah I see it .. So when you rearrange it... you do it according to the numbers and letters right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because you want the common terms to be together

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh okay, so for instance, if it said 6ab+6a, I would group that first? Does it matter which order it's in?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it depends... remember: in addition there is a commutative property...that means order doesn't matter... but in subtraction...there's no commutative property

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0unless you do it like this 6ab  6a => 6a + 6ab that's fine.. but these are not equal 6ab  6a => 6a  6ab

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh okay, so what would you do if it was in subtraction? Can you give me another question to solve to see if I understand?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sure try this \[\huge 5xy + 10y  4x^2 + 8y^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it's a bit different since you don't have same numbers here

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry...i have to go now...
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