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From my Algebra 2 class. Not homework.

$4x^3/2x^5y^2$

Divide the bases and subtract the exponents:

$2x^{-2}y^2$

Get rid of negative exponent by division:

$2y^2/x^2$

Then the answer should be:

$2x^2y^2$

Is this correct?

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    this website has a good explanation of some of the exponent rules. http://mathontrack.comze.com/exponentials2.html2014-10-09

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Here's how I think about the exponent rules. It may be helpful to you. Reduce the numerical part first and consider $ \frac{2x^3}{x^5y^2}. $ I notice that both have common factors of $x$. How many copies of $x$ are there? There are 3 on the top and 5 on the bottom, so I could think of the fraction as $ \frac{2xxx}{xxxxxy^2}. $ (Of course I would never actually write that down, but it's useful to keep in mind.) Now, if I went by and canceled factors of $x$ one-by-one, I would eventually be left with $ \frac{2}{xxy^2}, $ or better yet $ \frac{2}{x^2y^2}. $ The advantage of thinking in this way is I don't memorize unmotivated rules about when to add or subtract exponents (though, if you reflect for a moment, you'll see that this method is the same as the exponent rules you've learned) and avoids getting negative exponents unnecessarily.

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    This answer assumes that you intended the original problem to be $\frac{4x^3}{2x^5y^2}$ and not $\frac{4x^3}{2x^5} \cdot y^2.$ If you meant the latter, then it's all the same business except that $y^2$ will now be in the numerator.2011-09-15
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One of the problems with the "slant bar" notation is that it is not clear what fraction you have to begin with.

Does "$2x^2/x^5y^2$" represent $\frac{2x^2}{x^5y^2},$ or does it represent $\frac{2x^2}{x^5}\cdot y^2\quad?$

Normally, it would be interpreted as the first; but you seem to be interpreting it as the second. That would happen if, in the board or handwriting, there was a prominent and clear space, something like $2x^2/x^5\ \ y^2,$ which got lost along the way.

Nonetheless, because of this possibility of confusion, I strongly advice all my students to abandon the "slant bar" notation in mathematical formulas.

If your original problem was $\frac{2x^2}{x^5y^2}$ then your first step is wrong, because the $y^2$ is in the denominator. The final answer should be as Austin Mohr gives it.

If your original problem was $\frac{2x^2}{x^5}\cdot y^2,$ then everything you did was right up to the final step; you should not have gone from $\displaystyle\frac{2y^2}{x^2}$ to $\displaystyle 2y^2x^2$.