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anonymous

  • 5 years ago

I am a little lost on this 20x^8-40x^6+40x^2/5x^2

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  1. anonymous
    • 5 years ago
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    you can divide 5x^2 from both the top and the bottom. Divide each coefficient in the top by 5, and subtract 2 from each exponent. (x^m / x^n = x^(m-n))

  2. anonymous
    • 5 years ago
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    so would this be the answer 4x^6-8x^4+8

  3. anonymous
    • 5 years ago
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    yep!

  4. anonymous
    • 5 years ago
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    then how would you do this one 30x^3y^2/6x^7y^2

  5. anonymous
    • 5 years ago
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    would it be this 5X^10y^4

  6. anonymous
    • 5 years ago
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    Just do the same thing except for both x and y. You can do it in steps. First divide the top and bottom by 6. Then divide the top and bottom by x^7, then divide the top and bottom by y^2. When you are done you might want to the negative exponent to the denominator.

  7. anonymous
    • 5 years ago
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    No, you are close, but you are adding the exponents. You need to subtract when you are dividing.

  8. anonymous
    • 5 years ago
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    that would make them negative correct

  9. anonymous
    • 5 years ago
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    x^m / x^n = x^(m-n) Also, x^(-m) = 1/x^m

  10. anonymous
    • 5 years ago
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    Well, x^3/x^7 is x^(3-7) = x^(-4)

  11. anonymous
    • 5 years ago
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    well then I have stupid question how do you make it where the exponents are nonnegative

  12. anonymous
    • 5 years ago
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    x^(-m) = 1/x^m you can turn x^(-4) to 1/(x^4) A negative exponent just means that it is 1 over to what it would be if the exponent were positive. http://tutorial.math.lamar.edu/Classes/Alg/IntegerExponents.aspx

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