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

We can accomplish this division by re-writing the problem as a fraction.

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

The denominator will distribute, allowing us to address each element separately.

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

Now we can cancel common factors to find our answer.

$\displaystyle (\frac{18}{9}* \frac{x^{2}}{x})+1-(\frac{3}{9}* \frac{1}{x})$

$\displaystyle 2x+1-\frac{1}{3x}$

$\displaystyle y=\frac{(x^2-16)(x+7)}{x+4}+(x-1)^2+13$

First, factor the numerator of the quotient term by recognizing the difference of squares:

$\displaystyle y=\frac{(x-4)(x+4)(x+7)}{x+4}+(x-1)^2+13$

Cancel out the common term from the numerator and denominator:

$\displaystyle y=(x-4)(x+7)+(x-1)^2+13$

FOIL (First Outer Inner Last) the first two terms of the equation:

$\displaystyle y=x^2+7x-4x-28+x^2-2x+1+13$

Combine like terms:

$\displaystyle y=2x^2+x-14$

First, rewrite this problem so that the missing $\displaystyle x^{2}$ term is replaced by $\displaystyle 0x^{2}$

$\displaystyle (x^{3} + 0x^{2} -23x+10) \div (x-5)$

Divide the leading coefficients:

$\displaystyle x^{3} \div x = x^{2}$, the first term of the quotient

Multiply this term by the divisor, and subtract the product from the dividend:

$\displaystyle x^{2} (x-5) = x^{3} -5x^{2}$

$\displaystyle (x^{3} + 0x^{2} -23x+10) - (x^{3} -5x^{2}) = 5x^{2}-23x+10$

Repeat this process with each difference:

$\displaystyle 5x^{2} \div x = 5x$, the second term of the quotient

$\displaystyle 5x (x-5) = 5 x^{2} -25$

$\displaystyle 5x^{2}-23x+10 - ( 5x^{2} -25x) = 2x+10$

One more time:

$\displaystyle 2x \div x = 2$, the third term of the quotient

$\displaystyle 2 (x-5) = 2x-10$

$\displaystyle 2x+10 - (2x-10) = 20$, the remainder

The quotient is $\displaystyle x^{2} + 5x + 2$ and the remainder is $\displaystyle 20$; this can be rewritten as a quotient of 

$\displaystyle x^{2} + 5x + 2 + \frac{20}{x-5}$

Divide the leading coefficients to get the first term of the quotient:

$\displaystyle x^{2} \div x = x$, the first term of the quotient

Multiply this term by the divisor, and subtract the product from the dividend:

$\displaystyle x (x+3) = x^{2} + 3x$

$\displaystyle \left (x^{2} + 6x -7 \right ) - (x^{2}+3x) = 3x -7$

Repeat these steps with the differences until the difference is an integer. As it turns out, we need to repeat only once:

$\displaystyle 3x\div x = 3$, the second term of the quotient

$\displaystyle 3 (x+3) = 3x + 9$

$\displaystyle (3x -7) - (3x + 9) = -16$, the remainder

Putting it all together, the quotient can be written as $\displaystyle x+ 3 - \frac{16}{x+3}$.

Simplify the following expression:

$\displaystyle \frac{(x-25)}{(x^2-625)}$

To begin, we need to recognize the bottom as a difference of squares. Rewrite it as follows.

$\displaystyle \frac{(x-25)}{(x+25)(x-25)}=\frac{1}{x+25}$

So our answer is:

$\displaystyle \frac{1}{x+25}$

Simplify the following expression:

$\displaystyle \frac{(3x)(3x^2+5x)}{x}$

First, let's multiply the 3x through:

$\displaystyle \frac{(3x)(3x^2+5x)}{x}=\frac{9x^3+15x^2}{x}$

Next, divide out the x from the bottom:

$\displaystyle \frac{9x^3+15x^2}{x}=9x^2+15x$

So our answer is:

$\displaystyle 9x^2+15x$

$\displaystyle \begin{align*}&\textbf{To perform a polynomial long division, it is often}\\&\textbf{helpful to space terms into columns ordered by power.}\\&\textbf{This includes zero terms. For example: }x^2+0x+5\\&\textbf{When performing each division step, choose a coefficient}\\&\textbf{Which eliminates the highest ordered term each time.}\end{align*}\begin{align*}&&&&&&&&\end{align*}\begin{align*}&&&&\mathbf{x}&&\mathbf{+15}\\x&&+15|x^{2}&&+30x&&+225\\&&-(x^{2}&&+15x)\\&&&&15x&&+225\\&&&&-(15x&&+225)\\\end{align*}$

$\displaystyle \begin{align*}&\textbf{To perform a polynomial long division, it is often}\\&\textbf{helpful to space terms into columns ordered by power.}\\&\textbf{This includes zero terms. For example: }x^2+0x+5\\&\textbf{When performing each division step, choose a coefficient}\\&\textbf{Which eliminates the highest ordered term each time.}\end{align*}\begin{align*}&&&&&&&&\end{align*}\begin{align*}&&&&\mathbf{x}&&\mathbf{+6}\\x&&-5|x^{2}&&+x&&-30\\&&-(x^{2}&&-5x)\\&&&&6x&&-30\\&&&&-(6x&&-30)\\\end{align*}$

$\displaystyle \begin{align*}&\textbf{To perform a polynomial}\\&\textbf{long division, it is often}\\&\textbf{helpful to space terms}\\&\textbf{into columns ordered by power.}\\&\textbf{This includes zero terms.}\\&\textbf{For example: }x^2+0x+5\\&\textbf{When performing each}\\&\textbf{division step, choose a coefficient}\\&\textbf{Which eliminates the highest}\\&\textbf{ordered term each time.}\end{align*}\begin{align*}&&&&&&&&\end{align*}\begin{align*}&&&&\mathbf{x}&&\mathbf{+11}\\x&&+13|x^{2}&&+24x&&+143\\&&-(x^{2}&&+13x)\\&&&&11x&&+143\\&&&&-(11x&&+143)\\\end{align*}$

$\displaystyle \begin{align*}&\textbf{To perform a polynomial}\\&\textbf{long division, it is often}\\&\textbf{helpful to space terms}\\&\textbf{into columns ordered by power.}\\&\textbf{This includes zero terms.}\\&\textbf{For example: }x^2+0x+5\\&\textbf{When performing each}\\&\textbf{division step, choose a coefficient}\\&\textbf{Which eliminates the highest}\\&\textbf{ordered term each time.}\end{align*}\begin{align*}&&&&&&&&\end{align*}\begin{align*}&&&&&&&&\mathbf{x^{3}}&&\mathbf{+10x^{2}}&&\mathbf{-151x}&&\mathbf{-880}\\x^{2}&&-9x&&-112|x^{5}&&+x^{4}&&-353x^{3}&&-641x^{2}&&+24832x&&+98560\\&&&&-(x^{5}&&-9x^{4}&&-112x^{3})\\&&&&&&10x^{4}&&-241x^{3}&&-641x^{2}\\&&&&&&-(10x^{4}&&-90x^{3}&&-1120x^{2})\\&&&&&&&&-151x^{3}&&479x^{2}&&24832x\\&&&&&&&&-(-151x^{3}&&1359x^{2}&&16912x)\\&&&&&&&&&&-880x^{2}&&7920x&&98560\\&&&&&&&&&&-(-880x^{2}&&7920x&&98560)\\\end{align*}$