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anonymous

  • 4 years ago

How do I solve: 4y^2 - 11y - 3 =0 by factoring?

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  1. anonymous
    • 4 years ago
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    You should use quadratic formula for this.

  2. anonymous
    • 4 years ago
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    but I have to use factoring.

  3. anonymous
    • 4 years ago
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    |dw:1328452241616:dw|

  4. amistre64
    • 4 years ago
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    factoring means to undo multiplication

  5. anonymous
    • 4 years ago
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    okay

  6. amistre64
    • 4 years ago
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    we have to put this into a form that can be unproducted lol

  7. anonymous
    • 4 years ago
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    hm?

  8. amistre64
    • 4 years ago
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    a product is the result of multiplying; to undo multiplication means to factor it out again

  9. anonymous
    • 4 years ago
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    ohh. okay.

  10. amistre64
    • 4 years ago
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    3*2 produces 6 6 factors into 3 and 2

  11. anonymous
    • 4 years ago
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    Well I sort of factored but I have questions..

  12. anonymous
    • 4 years ago
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    4y^2 - 11y - 3 = 4y^2 + y - 12y - 3 = y(4y + 1) - 3(4y + 1) = (4y + 1)(y - 3)

  13. amistre64
    • 4 years ago
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    there are a few algorithms for factoring quadratics; pratu did factor by grouping there :)

  14. anonymous
    • 4 years ago
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    okay..

  15. anonymous
    • 4 years ago
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    how about this:

  16. anonymous
    • 4 years ago
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    But the point is: how does it help in solving the equation?

  17. anonymous
    • 4 years ago
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    4y^2 - 12y + y - 3=0 4y (y-3) + (y-3) = 0 (y-3) (4y+1) = 0

  18. amistre64
    • 4 years ago
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    same results :)

  19. anonymous
    • 4 years ago
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    but... I don't get why it's "(y-3)(4y + 1) =0

  20. amistre64
    • 4 years ago
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    undistributing

  21. anonymous
    • 4 years ago
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    like from 4y (y-3)+(y-3) to (y-3)(4y + 1) =0

  22. amistre64
    • 4 years ago
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    a(b+c) = ab + ac ab + ac = a(b+c)

  23. amistre64
    • 4 years ago
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    your fatoring out a (y-3) in this case

  24. anonymous
    • 4 years ago
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    ohhhh. factoring a y-3?

  25. amistre64
    • 4 years ago
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    yep

  26. anonymous
    • 4 years ago
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    what do you mean?

  27. amistre64
    • 4 years ago
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    grouping allows you to "see" the common factor of (y-3) right?

  28. anonymous
    • 4 years ago
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    yes

  29. amistre64
    • 4 years ago
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    then factor it out ....

  30. anonymous
    • 4 years ago
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    (4y + 1)(y - 3) = 0 |dw:1328452809130:dw|

  31. anonymous
    • 4 years ago
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    ohhh.. I think I'm getting it, ugh. Why is math so difficult for me >_<

  32. amistre64
    • 4 years ago
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    you have 2 terms: 4y (y-3) and 1(y-3) the GCF is (y-3)

  33. anonymous
    • 4 years ago
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    ohh, it is.

  34. amistre64
    • 4 years ago
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    4(2) + 1(2) is the same as 2(4+1) right?

  35. anonymous
    • 4 years ago
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    yes

  36. anonymous
    • 4 years ago
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    but why is it (y-3)(4y+1) now? o.o

  37. amistre64
    • 4 years ago
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    well, in this case it aint a 2, its (y-3)

  38. anonymous
    • 4 years ago
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    |dw:1328452856397:dw|

  39. anonymous
    • 4 years ago
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    = (-y + 4)/(y - 3)

  40. anonymous
    • 4 years ago
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    Does it help in any way?

  41. anonymous
    • 4 years ago
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    4y = (-y + 4)/(y - 3)

  42. anonymous
    • 4 years ago
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    yes it does

  43. anonymous
    • 4 years ago
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    thank you guys!

  44. amistre64
    • 4 years ago
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    yw

  45. anonymous
    • 4 years ago
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    yw

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