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anonymous

  • one year ago

Simplify the rational expression. State any restrictions on the variable.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Nnesha do you know anything about this?

  3. anonymous
    • one year ago
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    @jim_thompson5910

  4. jim_thompson5910
    • one year ago
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    what two numbers multiply to 24 and add to -10?

  5. anonymous
    • one year ago
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    12 and 2

  6. jim_thompson5910
    • one year ago
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    nope

  7. jim_thompson5910
    • one year ago
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    12+2 = 14 not -10

  8. anonymous
    • one year ago
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    |dw:1444458852967:dw| I believe this is the answer

  9. anonymous
    • one year ago
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    opps for some reason I thought i read multiply lol thats what I get for 15 hours of school work today

  10. jim_thompson5910
    • one year ago
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    -6 and -4 fit the description, no? -6 + (-4) = -10 -6 times -4 = 24

  11. anonymous
    • one year ago
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    Yes that is correct. I see what you mean by that now

  12. jim_thompson5910
    • one year ago
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    so that means n^4 - 10n^2 + 24 factors to (n^2 - 6)(n^2 - 4) we can then further factor n^2 - 4 down into (n-2)(n+2)

  13. jim_thompson5910
    • one year ago
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    so overall, n^4 - 10n^2 + 24 completely factors to (n^2 - 6)(n-2)(n+2)

  14. jim_thompson5910
    • one year ago
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    you'll do the same for the denominator and see if you can cancel anything

  15. anonymous
    • one year ago
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    Okay so the bottom line reads n^4-9n^2+18 Im confused:/ Would I need to find a number that multiplies to 24 and adds to -9?

  16. jim_thompson5910
    • one year ago
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    multiplies to 18 actually

  17. jim_thompson5910
    • one year ago
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    find two numbers that multiply to 18 and add to -9

  18. anonymous
    • one year ago
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    so 6 and 3 right cause 6*3=18 and -6+-3=-9

  19. jim_thompson5910
    • one year ago
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    -6 and -3

  20. jim_thompson5910
    • one year ago
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    I think that's what you meant to say?

  21. anonymous
    • one year ago
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    right

  22. jim_thompson5910
    • one year ago
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    so n^4-9n^2+18 factors to (n^2-6)(n^2-3)

  23. jim_thompson5910
    • one year ago
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    n^4 - 10n^2 + 24 completely factors to (n^2 - 6)(n-2)(n+2) n^4-9n^2+18 completely factors to (n^2-6)(n^2-3)

  24. jim_thompson5910
    • one year ago
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    at this point, I see a pair of terms cancelling

  25. anonymous
    • one year ago
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    n^2-6

  26. jim_thompson5910
    • one year ago
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    yes

  27. anonymous
    • one year ago
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    so (n^2-2) (n+2) (n^2-3) (n^2-4) (n^2-3)

  28. jim_thompson5910
    • one year ago
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    I think you meant to say `(n-2) (n+2)` in your first step

  29. jim_thompson5910
    • one year ago
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    but yeah the final answer is \[\Large \frac{n^2-4}{n^2-3}\]

  30. jim_thompson5910
    • one year ago
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    what would the restrictions be?

  31. anonymous
    • one year ago
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    so would the answer be this |dw:1444460035990:dw|

  32. jim_thompson5910
    • one year ago
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    close

  33. jim_thompson5910
    • one year ago
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    but no

  34. anonymous
    • one year ago
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    could we work the problem out further? I quess i still dont 100% understand

  35. jim_thompson5910
    • one year ago
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    the denominator factors into (n^2-6)(n^2-3) each piece is set equal to 0 and solved for to get n^2-6 = 0 n^2 = 6 n = sqrt(6) or n = -sqrt(6) n^2-3 = 0 n^2 = 3 n = sqrt(3) or n = -sqrt(3)

  36. anonymous
    • one year ago
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    would it be the square root of those numbers ( 6, -3)

  37. jim_thompson5910
    • one year ago
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    so the restrictions are actually \[\Large n \ne \pm \sqrt{6}\] \[\Large n \ne \pm \sqrt{3}\]

  38. anonymous
    • one year ago
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    |dw:1444460209535:dw|

  39. jim_thompson5910
    • one year ago
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    much better

  40. anonymous
    • one year ago
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    thank you so much!! You are a lifesaver and such a great teacher! My husband usually helps me with math but he doesn't know this type of stuff! Im giving you a medal for your help.

  41. jim_thompson5910
    • one year ago
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    I'm glad it's making more sense now

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