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anonymous
 3 years ago
Factor each completely: 14v^32v^235v+5
anonymous
 3 years ago
Factor each completely: 14v^32v^235v+5

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Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1368509642140:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Luigi0210 Why in the last senetence is the 5 a negative instead of a positive?

mathstudent55
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1368510000536:dw

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.2Well you have to factor out a 5 from the second part to get the equations to match

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh! It's because I did it like this: 5(7v+1) and I still got the same answer

mathstudent55
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1368510556848:dw

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1368510657852:dw If you got that when you factored then it would be wrong because the two equations are not the same. When factoring by grouping, equations have to be EXACTLY the same for it to work.

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.2Unless you factored out a 2v^2..

mathstudent55
 3 years ago
Best ResponseYou've already chosen the best response.1Your factoring is correct in the sense that what you got is equal to the unfactored expression, but since there is no common factor you can't continue. The goal of factoring by groups is to get a common factor.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, I'm getting what you guys are saying now. My teacher said though that I am able to change the sign around to get it to match

mathstudent55
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1368510916740:dw

mathstudent55
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1368511042206:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@mathstudent55 AHHH!!!! I SEE NOW!! There's always going to be a way for it to equal right?

Luigi0210
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1368511111775:dw

mathstudent55
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1368511115956:dw

mathstudent55
 3 years ago
Best ResponseYou've already chosen the best response.1If you are asked specifically to factor a polynomial by grouping, then chances are you are given a problem that can be factored by grouping. In this case, there will be a common factor that you can pull out. You just may have to adjust the sign of what you pull out to make sure what's left is equal to what was left in the first group.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@mathstudent55 @Luigi0210 Yeah, this makes alot of sence now thank you so much:D
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