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:  \(\displaystyle \frac{x-30}{x-5}\)

Write an expression to divide the polynomials.

\(\displaystyle \frac{x^2-2x+1}{2x^2-2}\)

Both polynomials can be factorized.  

The numerator will be simplified to \(\displaystyle (x-1)^2\).

The denominator cannot be factorized as is.  Pull a common factor of two in the denominator.

\(\displaystyle 2x^2-2 = 2(x^2-1)\)

The \(\displaystyle x^2-1\) term can be factorized to \(\displaystyle (x+1)(x-1)\).

We can now rewrite the fraction.

\(\displaystyle \frac{x^2-2x+1}{2x^2-2} = \frac{(x-1)^2}{2(x+1)(x-1)}\)

The common \(\displaystyle (x-1)\) terms in the numerator and denominator can be cancelled.

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

Distribute the two in the denominator through both terms in the binomial.

The answer is:  \(\displaystyle \frac{x-1}{2x+2}\)