## ineedyouubiebs 2 years ago Factor each completely: 14v^3-2v^2-35v+5

1. Luigi0210

|dw:1368509642140:dw|

2. ineedyouubiebs

@Luigi0210 Why in the last senetence is the 5 a negative instead of a positive?

3. mathstudent55

|dw:1368510000536:dw|

4. Luigi0210

Well you have to factor out a -5 from the second part to get the equations to match

5. ineedyouubiebs

Oh! It's because I did it like this: 5(-7v+1) and I still got the same answer

6. ineedyouubiebs

???

7. mathstudent55

|dw:1368510556848:dw|

8. ineedyouubiebs

@mathstudent55 Yeah

9. Luigi0210

|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.

10. Luigi0210

Unless you factored out a -2v^2..

11. mathstudent55

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.

12. ineedyouubiebs

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

13. mathstudent55

|dw:1368510916740:dw|

14. mathstudent55

|dw:1368511042206:dw|

15. ineedyouubiebs

@mathstudent55 AHHH!!!! I SEE NOW!! There's always going to be a way for it to equal right?

16. Luigi0210

|dw:1368511111775:dw|

17. mathstudent55

|dw:1368511115956:dw|

18. mathstudent55

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.

19. ineedyouubiebs

@mathstudent55 @Luigi0210 Yeah, this makes alot of sence now thank you so much:D