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ineedyouubiebs Group Title

Factor each completely: 14v^3-2v^2-35v+5

  • one year ago
  • one year ago

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  1. Luigi0210 Group Title
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    |dw:1368509642140:dw|

    • one year ago
  2. ineedyouubiebs Group Title
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    @Luigi0210 Why in the last senetence is the 5 a negative instead of a positive?

    • one year ago
  3. mathstudent55 Group Title
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    |dw:1368510000536:dw|

    • one year ago
  4. Luigi0210 Group Title
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    Well you have to factor out a -5 from the second part to get the equations to match

    • one year ago
  5. ineedyouubiebs Group Title
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    Oh! It's because I did it like this: 5(-7v+1) and I still got the same answer

    • one year ago
  6. ineedyouubiebs Group Title
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    ???

    • one year ago
  7. mathstudent55 Group Title
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    |dw:1368510556848:dw|

    • one year ago
  8. ineedyouubiebs Group Title
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    @mathstudent55 Yeah

    • one year ago
  9. Luigi0210 Group Title
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    |dw: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.

    • one year ago
  10. Luigi0210 Group Title
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    Unless you factored out a -2v^2..

    • one year ago
  11. mathstudent55 Group Title
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    Your 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.

    • one year ago
  12. ineedyouubiebs Group Title
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    Yeah, 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

    • one year ago
  13. mathstudent55 Group Title
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    |dw:1368510916740:dw|

    • one year ago
  14. mathstudent55 Group Title
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    |dw:1368511042206:dw|

    • one year ago
  15. ineedyouubiebs Group Title
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    @mathstudent55 AHHH!!!! I SEE NOW!! There's always going to be a way for it to equal right?

    • one year ago
  16. Luigi0210 Group Title
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    |dw:1368511111775:dw|

    • one year ago
  17. mathstudent55 Group Title
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    |dw:1368511115956:dw|

    • one year ago
  18. mathstudent55 Group Title
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    If 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.

    • one year ago
  19. ineedyouubiebs Group Title
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    @mathstudent55 @Luigi0210 Yeah, this makes alot of sence now thank you so much:D

    • one year ago
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