anonymous
  • anonymous
I am a little lost on this 20x^8-40x^6+40x^2/5x^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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))
anonymous
  • anonymous
so would this be the answer 4x^6-8x^4+8
anonymous
  • anonymous
yep!

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anonymous
  • anonymous
then how would you do this one 30x^3y^2/6x^7y^2
anonymous
  • anonymous
would it be this 5X^10y^4
anonymous
  • anonymous
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.
anonymous
  • anonymous
No, you are close, but you are adding the exponents. You need to subtract when you are dividing.
anonymous
  • anonymous
that would make them negative correct
anonymous
  • anonymous
x^m / x^n = x^(m-n) Also, x^(-m) = 1/x^m
anonymous
  • anonymous
Well, x^3/x^7 is x^(3-7) = x^(-4)
anonymous
  • anonymous
well then I have stupid question how do you make it where the exponents are nonnegative
anonymous
  • anonymous
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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