\(\displaystyle \frac{x^2-10x+25}{x^2-25}\)

First we will factor the numerator:

\(\displaystyle x^2-10x+25=(x-5)(x-5)\)

Then factor the denominator:

\(\displaystyle x^2-25=(x+5)(x-5)\)

We can re-write the original fraction with these factors and then cancel an (x-5) term from both parts:

\(\displaystyle \frac{(x-5)(x-5)}{(x-5)(x+5)}=\frac{x-5}{x+5}\)

First, set up the division as the following:

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \end{array}\)

Look at the leading term \(\displaystyle x^2\) in the divisor and \(\displaystyle 3x^3\) in the dividend. Divide \(\displaystyle 3x^3\) by \(\displaystyle x^2\) gives \(\displaystyle 3x\); therefore, put \(\displaystyle 3x\) on the top:

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \end{array}\)

Then take that \(\displaystyle 3x\) and multiply it by the divisor, \(\displaystyle x^2-1\), to get \(\displaystyle 3x^3-3x\).  Place that \(\displaystyle 3x^3-3x\) under the division sign:

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \end{array}\)

Subtract the dividend by that same \(\displaystyle 3x^3-3x\) and place the result at the bottom. The new result is \(\displaystyle 5x^2+2x+8\), which is the new dividend.

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \end{array}\)

Now, \(\displaystyle 5x^2\) is the new leading term of the dividend.  Dividing \(\displaystyle 5x^2\) by \(\displaystyle x^2\) gives 5.  Therefore, put 5 on top:

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & 5 & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \end{array}\)

Multiply that 5 by the divisor and place the result, \(\displaystyle 5x^2-5\), at the bottom:

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & 5 & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \\ & && & 5x^2 & -5 \end{array}\)

Perform the usual subtraction:

\(\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & 5 & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \\ & && & 5x^2 & -5 \\ \cline{5-7} & && & & 2x & 13 \\ \end{array}\)

Therefore the answer is \(\displaystyle 3x+5\) with a remainder of \(\displaystyle 2x+13\), or \(\displaystyle 3x+5+\frac{2x+13}{x^2-1}\).

When dividing polynomials, subtract the exponent of the variable in the numberator by the exponent of the same variable in the denominator.

If the power is negative, move the variable to the denominator instead.

First move the negative power in the numerator to the denominator:

\(\displaystyle \frac{x^{2}y^{15}}{x^{5}y^{4}z^{2}}\)

Then subtract the powers of the like variables:

\(\displaystyle \frac{y^{11}}{x^{3}z^{2}}\)

The numerator is equivalent to

\(\displaystyle \frac{x-5+2x}{x-5}\)\(\displaystyle =\frac{3x-5}{x-5}\)

The denominator is equivalent to

\(\displaystyle \frac{x-5+3}{x-5}\) \(\displaystyle =\frac{x-2}{x-5}\)

Dividing the numerator by the denominator, one gets

\(\displaystyle \frac{3x-5}{x-2}\)

First let us find a common denominator as follows:

\(\displaystyle \frac{2x\left ( x+1 \right )}{\left ( x+1 \right )\left ( x-1 \right )} - \frac{3}{\left ( x+1 \right )\left ( x-1 \right )}\)

Now we can subtract the numerators which gives us : \(\displaystyle 2x^{2}+2x-3\)

So the final answer is \(\displaystyle \frac{2x^{2}+2x-3}{\left ( x+1 \right )\left ( x-1 \right )}\)

Factor both the numerator and the denominator which gives us the following:

\(\displaystyle \frac{\left ( 2x-3 \right )\left ( 3x+1 \right )}{\left ( 2x-3 \right )\left ( 3x+5 \right )}\)

After cancelling \(\displaystyle \left ( 2x-3 \right )\) we get

\(\displaystyle \frac{3x+1}{3x+5}\)

We are dividing the polynomial by a monomial. In essence, we are dividing each term of the polynomial by the monomial. First I like to re-write this expression as a fraction. So,

\(\displaystyle (24x^8-2x^6+12x^2)\div x^2 = \frac{24x^8-2x^6+12x^2}{x^2}\)

So now we see the three terms to be divided on top. We will divide each term by the monomial on the bottom. To show this better, we can rewrite the equation. \(\displaystyle \frac{24x^8-2x^6+12x^2}{x^2} = \frac{24x^8}{x^2}-\frac{2x^6}{x^2}+\frac{12x^2}{x^2}\)

Now we must remember the rule for dividing variable exponents. The rule is \(\displaystyle \frac{x^a}{x^b} = x^{a-b}\)So, we can use this rule and apply it to our expression above. Then,\(\displaystyle \frac{24x^8}{x^2}-\frac{2x^6}{x^2}+\frac{12x^2}{x^2} = 24x^{8-2}-2x^{6-2}+2x^{2-2}=24x^6-2x^4+12\)

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 \(\displaystyle x+ 3 - \frac{16}{x+3}\).

In order to divide these polynomials, we will need to factorize both the top and the bottom expressions.

\(\displaystyle \frac{x^2-28x-60}{x^2-3x-10} = \frac{(x-30)(x+2)}{(x-5)(x+2)}\)

Cancel out the common terms in the numerator and denominator.

The answer is:  \(\displaystyle \frac{x-30}{x-5}\)