When dividing exponents, you subtract exponents that share the same base, so

$\displaystyle 5-8=-3$ and $\displaystyle 10-(-6)=16$ and $\displaystyle 3-2=1$.

Do not forget to "add the opposite" when subtracting negative numbers).

Now, you have

$\displaystyle x^{-3}y^{16}z^{1}$

But you are not done yet!  Remember, you do not want to have a negative exponent, and the way to turn the negative exponent into a positive exponent is to take its reciprocal, like this:

$\displaystyle x^{-3} = \frac{1}{x^{3}}$

You keep the rest of the equation in the numerator, leaving you with

$\displaystyle \frac{y^{16}z}{x^{3}}$

$\displaystyle \frac{r^2s^9x^5}{r^4s^3x}$

To simplify, we must use exponent rules. For exponents in fractions, we can subtract the exponent of the denominator from the exponent in the numerator.

$\displaystyle \frac{a^b}{a^c}=a^{b-c}$

With this rule, we can rewrite the problem.

$\displaystyle r^{2-4}s^{9-3}x^{5-1}$

$\displaystyle r^{-2}s^6x^4$

Remember that negative exponents get moved back to the denominator, turning them positive.

$\displaystyle \frac{s^6x^4}{r^2}$

4 goes into 24, 12, 8, and 4.

Similarly, the smallest exponent of x in the four terms is 2, and the smallest exponent of y in the four terms is 1.

Hence the GCF must be $\displaystyle 4x^{2}y$.

Divide each of the terms in the numerator by the denominator:

$\displaystyle \frac{4x^{2}y^{2}}{4x^{2}y}-\frac{8xy^{3}}{4x^{2}y} + \frac{16y^{4}}{4x^{2}y}$

Simplify each term above to get the final:

$\displaystyle y-\frac{2y^{2}}{x}+\frac{4y^{3}}{x}$

The numerator can be factored into

$\displaystyle \left ( x-1 \right )\left ( x+3 \right )\left ( 2x-3 \right )$,

which when divided by $\displaystyle \left ( x+3 \right )$,

gives us $\displaystyle \left ( x-1 \right )\left ( 2x-3 \right ) = 2x^{2}-5x+3$.

Alternate method: Long division of the numerator by the denominator gives the same answer.

When we divide a polynomial by another polynomial we get:

1. Quotient
2. Remainder (if one exists)

In our problem the long division results in:

1. A quotient of $\displaystyle x^{2}+2x-3$
2. A remainder of $\displaystyle \frac{-6}{2x-3}$

This can easily be solved by factoring using the difference of cubes formula:

$\displaystyle \left ( x^{3}-y^{3} \right ) = \left ( x-y \right )\left ( x^{2}+xy+y^{2} \right )$

First, convert the given polynomial into a difference of two cubes:

$\displaystyle \left ( \left ( 2x \right )^{3} -\left ( 1 \right )^{3}\right )$

Compare this with the difference of cubes formula above to get:

$\displaystyle \left ( 2x-1 \right )\left ( 4x^{2}+2x+1 \right )$

By dividing the above numerator by the given denominator we get:

$\displaystyle 4x^{2}+2x+1$

$\displaystyle (5t^{3} - 40t^{2}+ 30t- 10) \div 10t$ 

$\displaystyle = \frac{5t^{3}}{10t} - \frac{40t^{2}}{10t}+ \frac{30t}{10t}- \frac{10}{10t}$

Cancel:

$\displaystyle = \frac{t^{3-1}}{2} - \frac{4t^{2-1}}{1}+ \frac{3}{1}- \frac{1}{t}$

$\displaystyle = \frac{1}{2} t^{2}- 4t+ 3- \frac{1}{t}$

$\displaystyle \left ( 9x ^{2} + 21x - 18\right ) \div \left (-3x \right )$

$\displaystyle =\frac{9x ^{2}}{-3x } +\frac{21x }{-3x }- \frac{18}{-3x }$

$\displaystyle =-\frac{9x ^{2}}{3x }-\frac{21x }{3x }+ \frac{18}{3x }$

Cancel:

$\displaystyle =-\frac{3x ^{2-1}}{1 }-\frac{7 }{1 }+ \frac{6}{x }$

$\displaystyle =-3x -7+\frac{6}{x }$

7 in the denominator is a common factor of the three coefficients in the numerator, which allows you to divide out the 7 from the denominator:

$\displaystyle \frac {3b^2 + b^3 - 2b}{b}$

Then divide by $\displaystyle b$:

$\displaystyle ={3b + b^2 - 2}$