3
$\begingroup$

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?

  • 0
    assuming you mean $\frac{4x^3y^2}{2x^5}$ then you your first step is valid as is your second but note that "getting rid of the negative exponent by division" what you are really doing here is just writing out explicity what $x^{-2}$ represents. The final step however is not correct as $\frac{2y^2}{x^2} \neq 2x^2y^2$2011-09-15
  • 0
    It's what I was told to do. I figured it was wrong anyway2011-09-15
  • 1
    @david: If the original fraction is $\Bigl((4x^3)/(2x^5)\bigr)y^2$, then $(2y^2)/(x^2)$ is correct. But if the original fraction was $$\frac{4x^3}{2x^5y^2},$$ then the correct final answer should be $2x^{-2}y^{-2}$.2011-09-15
  • 0
    Note that fraction slashes often lead to ambiguity, as here. Please either use parentheses or stack the fraction vertically using \frac{numerator}{denominator}2011-09-15
  • 0
    this website has a good explanation of some of the exponent rules. http://mathontrack.comze.com/exponentials2.html2014-10-09

2 Answers 2