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lalaland_lauren

  • one year ago

Determine the factors of 15x2 + 3xy + 10x + 2y. (3x + 2)(5x + y) (5x + y)(2x + 3) (3x + y)(5x + 2) (5x + 3)(2x + y)

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  1. lalaland_lauren
    • one year ago
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    @satellite73

  2. lalaland_lauren
    • one year ago
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    @vmac2122

  3. misty1212
    • one year ago
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    HI!!

  4. lalaland_lauren
    • one year ago
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    Hello :)

  5. misty1212
    • one year ago
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    we can multiply each one out and see what works, or perhaps we can work smarter

  6. misty1212
    • one year ago
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    you see how the last term is \(2y\) right?

  7. lalaland_lauren
    • one year ago
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    That's my problem, I get so confused multiplying everything out :( and yeah

  8. misty1212
    • one year ago
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    that gets rid if b and d do you see why?

  9. lalaland_lauren
    • one year ago
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    Yeah, that's actually a really cool loophole

  10. misty1212
    • one year ago
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    \[ (5x + y)(2x + 3)\] last term will be \(3y\) so that is no good

  11. misty1212
    • one year ago
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    same here \[ (5x + 3)(2x + y)\]

  12. lalaland_lauren
    • one year ago
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    I figured it out, it's A

  13. misty1212
    • one year ago
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    so we can try the first one \[ (3x + 2)(5x + y)\]

  14. misty1212
    • one year ago
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    yeah that looks good

  15. lalaland_lauren
    • one year ago
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    Yayyyyy I got it! :)

  16. misty1212
    • one year ago
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    \[\huge \color\magenta\heartsuit\]

  17. lalaland_lauren
    • one year ago
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    Thank youuuu!

  18. misty1212
    • one year ago
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    you are quite welcome got more?

  19. lalaland_lauren
    • one year ago
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    A few, yes hold on

  20. misty1212
    • one year ago
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    kk

  21. lalaland_lauren
    • one year ago
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    Graph of a cubic polynomial that falls to the left and rises to the right with graph touching x axis at negative 1 and x intercept at 1 Which of the following functions best represents the graph? f(x) = x3 + x2 − 4x − 4 f(x) = x3 + x2 − x − 1 f(x) = x3 + 3x2 − 4x − 12 f(x) = x3 + 2x2 − 2x − 1

  22. lalaland_lauren
    • one year ago
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  23. misty1212
    • one year ago
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    ok we can do this

  24. misty1212
    • one year ago
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    you see how it only touches where \(x=-1\)?

  25. lalaland_lauren
    • one year ago
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    Yeah, and where x=1

  26. misty1212
    • one year ago
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    actually there is a difference, where \(x=1\) it CROSSES the axis, at \(x=-1\) it comes up, touches it, but does not cross, goes back down

  27. misty1212
    • one year ago
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    so it has to zeros, one at \(x=-1\) and one at \(x=1\) because it crosses at \(x=1\) one factor is \((x-1)\)

  28. misty1212
    • one year ago
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    because it touches at \(x=-1\) then the factor looks like \((x+1)^2\)

  29. misty1212
    • one year ago
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    lol "has two zeros" i meant

  30. misty1212
    • one year ago
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    final job is to multiply \[(x+1)^2(x-1)\] to see what we get i would cheat!!

  31. misty1212
    • one year ago
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    http://www.wolframalpha.com/input/?i=%28x%2B1%29%5E2%28x-1%29

  32. misty1212
    • one year ago
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    looks like \[x^3+x^2-x-1\] is a winner is that a choice?

  33. lalaland_lauren
    • one year ago
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    Yup :) Wolfram Alpha has been helping me a lot this unit lol

  34. misty1212
    • one year ago
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    might as well use it instead of making an algebra mistake i do!

  35. lalaland_lauren
    • one year ago
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    Haha! I have just one more question and them I'm done lol

  36. lalaland_lauren
    • one year ago
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    Which shows 602 − 402 being evaluated using the difference of squares method? 602 − 402 = (3600 + 1600)(3600 − 1600) = 10,400,000 602 − 402 = 3600 − 1600 = 2,000 602 − 402 = (60 − 40)2 = 202 = 400 602 − 402 = (60 + 40)(60 − 40) = (100)(20) = 2,000

  37. lalaland_lauren
    • one year ago
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    The twos are squares

  38. misty1212
    • one year ago
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    \[60^2-40^2\]?

  39. lalaland_lauren
    • one year ago
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    Wait nvm I found the answer haha

  40. misty1212
    • one year ago
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    D right?

  41. lalaland_lauren
    • one year ago
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    Yup xD

  42. lalaland_lauren
    • one year ago
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    I got an 88% :D

  43. misty1212
    • one year ago
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    that it?

  44. misty1212
    • one year ago
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    then time for a tune https://www.youtube.com/watch?v=smdNetoPFV0

  45. lalaland_lauren
    • one year ago
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    I LOVE THAT SONG!!!!!

  46. lalaland_lauren
    • one year ago
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    I love that whole band, actually!

  47. misty1212
    • one year ago
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    lol thought so !!

  48. lalaland_lauren
    • one year ago
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    How'd you know? XD

  49. misty1212
    • one year ago
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    american psychic

  50. lalaland_lauren
    • one year ago
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    *clapping emoji*

  51. misty1212
    • one year ago
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    \[\color\red\heartsuit\]

  52. lalaland_lauren
    • one year ago
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    Unfortunately I have to go to bed now :( I'm extremely tired and I'm pretty sure I have to get up early to keep working on assignments. Will you be on in the morning?

  53. misty1212
    • one year ago
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    i can be here at about 9 or 9:30 earlier probably not

  54. misty1212
    • one year ago
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    tag me if you like, i will try to help if i can!

  55. misty1212
    • one year ago
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    my bed time too

  56. lalaland_lauren
    • one year ago
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    That's a good time for me :) alrighty, thanks so much for helping me! good night c:

  57. misty1212
    • one year ago
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    gnight c:

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