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anonymous

  • 5 years ago

I am sure there is someone here that can easily tackle this problem! Factor: 3x^(-1/2)+9x^91/2)-81^(-3/2) I panic when I see negative fractional exponents! And my factoring skills are weak at best If you can explain any of the steps or the way you think about the problem I would greatly appreciate the advice.

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  1. anonymous
    • 5 years ago
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    \[ 3x^{\frac{-1}{2}}+9x^{\frac{1}{2}}-81^{\frac{-3}{2}}\] like this?

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    I was thinking that an x^(1/2) If 81x^(-3/2) could be rewritten as 81x(-3(1/2)) and 3x(-1/2) could be rewritten as 3x^(-1(1/2))

  4. anonymous
    • 5 years ago
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    i can write out what this means without negative exponents, but i cannot see how to factor it

  5. anonymous
    • 5 years ago
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    oops that an x^(1/2) could be factored out *

  6. anonymous
    • 5 years ago
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    oh wait, is there an x in the last term?

  7. anonymous
    • 5 years ago
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    yes there was

  8. anonymous
    • 5 years ago
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    \[3x^{\frac{-1}{2}}+9x^{\frac{1}{2}}-81x^{\frac{-3}{2}}\]

  9. anonymous
    • 5 years ago
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    oh yikes I overlooked that

  10. anonymous
    • 5 years ago
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    good catch. I looked over it so quickly the first time I missed the omission.

  11. anonymous
    • 5 years ago
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    ok that is a whole different story. first off you can factor a 3 out of each term, then also perhaps an \[x^{\frac{1}{2}}\] if you like. lets try it

  12. anonymous
    • 5 years ago
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    Great I get coefficients of 1 3 and 27

  13. anonymous
    • 5 years ago
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    \[3x^{\frac{1}{2}}(x^{-1}+3+27x^{-2})\]

  14. anonymous
    • 5 years ago
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    now i wonder if we can go further...

  15. anonymous
    • 5 years ago
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    cool that exactly where I am at the moment as well

  16. anonymous
    • 5 years ago
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    oh I had -27^(-3)

  17. anonymous
    • 5 years ago
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    no i think it is \[27x^{-2}\]

  18. anonymous
    • 5 years ago
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    because \[\frac{-3}{2}-\frac{1}{2}=-2\]

  19. anonymous
    • 5 years ago
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    i cannot see how to factor the second part, so i think you are done at that step

  20. anonymous
    • 5 years ago
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    ok so the exponents are not being multiplied in this case

  21. anonymous
    • 5 years ago
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    I was thinking it was -27x^(-3(1/2))

  22. anonymous
    • 5 years ago
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    sorry i was off by a minus sign, should be \[-27x^{-2}\]

  23. anonymous
    • 5 years ago
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    when you multiply you add the exponents

  24. anonymous
    • 5 years ago
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    so \[3x^{\frac{1}{2}}\times -27x^{-2}=-81x^{\frac{1}{2}+(-2)}=-81x^{-\frac{3}{2}}\] which is what you want

  25. anonymous
    • 5 years ago
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    ok I think I see how that works so x^(1/2)x^-2(2/2) = x^(1/2)x^(-4/4) or x^(-3/2)

  26. anonymous
    • 5 years ago
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    Thank you so much for your help! This really helped clarify the concept for me.

  27. anonymous
    • 5 years ago
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    you don't need to change the denominator is 2

  28. anonymous
    • 5 years ago
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    yw

  29. anonymous
    • 5 years ago
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    oops meant (-2(2/2) = (-4/2)

  30. anonymous
    • 5 years ago
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    got it

  31. anonymous
    • 5 years ago
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    awesome!

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