anonymous
  • anonymous
I am a little lost on this 20x^8-40x^6+40x^2/5x^2
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.