Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.

Remember to place a \(\displaystyle 0\) when there isn't a coefficient given.

We multiply what's below the line by \(\displaystyle -1\) and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

\(\displaystyle 1x^{5}-1x^{4}+1x^{3}-1x^{2}+0x+0\) with remainder \(\displaystyle -3\)

This can be rewritten as:

\(\displaystyle x^{5}-x^{4}+x^{3}-x^{2}-\frac{3}{x+1}\)
Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.

To divide synthetically, we begin by drawing a box. On the inside separated by spaces, we write the coefficients of the terms of our polynomial being divided. On the outside, we write the root that would satisfy our binomial \(\displaystyle x-3\), namely \(\displaystyle 3\).  Leaving a space for another row of numbers, we then draw a line below our row of coefficients.

We then begin dividing by simply carrying our first coefficient (1) down below the line.

We then multiply this 1 by our divisor (3) and write the resulting product (3) below our next coefficient.

We then add the two numbers in that column and write the sum (5) below the line.

We then simply continue the process by multiplying this 5 by our divisor 3 and writing that product in the next column, adding it to the next coefficient, and continuing until we finish the columns.

We then need to translate our bottom row of numbers into the coefficients of our new quotient. Since the first column originally corresponded to our cubic term, it will now correspond to the quadratic term meaning that our 1 can be translated as \(\displaystyle x^2\).  Similarly, our second column transitions from quadratic to linear, making our 5 become \(\displaystyle 5x\).  Finally, our third column becomes the constant term, meaning 8 simply remains the constant 8.  Finally, our former constant column becomes the column for our remainder.  However, since we have a 0, we have no remainder and can disregard it.

Putting all of this together gives us a final answer of 

\(\displaystyle x^2+5x+8\)

First, set up the synthetic division problem by lining up the coefficients. There are a couple of different strategies - for this one, we will put a -7 in the top corner and add the columns.

\(\displaystyle \underline{-7}| \quad 1 \quad 1 \quad -36 \quad 42\)

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The first step is to bring down the first 1. Then multiply what is below the line by the -7 in the box, write it below the next coefficient, and then add the columns:

\(\displaystyle \underline{-7}| \quad 1 \quad 1 \quad -36 \quad 42\)

                  \(\displaystyle -7 \quad 42 \quad -42\)

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            \(\displaystyle 1 \quad -6 \quad 6 \quad 0\)

We can interpret this answer as meaning \(\displaystyle m^2 -6m +6\)

Our first step is to list the coefficiens of the polynomials in descending order and carry down the first coefficient.

We multiply what's below the line by \(\displaystyle 2\) and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficient.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

\(\displaystyle 1x^{2}+0x+4\) with reminder \(\displaystyle 5\)

This can be rewritten as:

\(\displaystyle x^{2}+4+\frac{5}{x-2}\)

Factorize the numerator for the function: 

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

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

The removable discontinuity is \(\displaystyle x-1\) since this is a term that can be eliminated from the function.  There are no vertical asymptotes.

Set the removable discontinutity to zero and solve for the location of the hole.

 \(\displaystyle x-1=0\)

\(\displaystyle x=1\)

The hole is located at:  \(\displaystyle x=1\)

Rewrite the function \(\displaystyle y=\frac{x^2-6x+9}{x^2-9}\) in its factored form.

\(\displaystyle y=\frac{x^2-6x+9}{x^2-9}=\frac{(x-3)(x-3)}{(x+3)(x-3)}\)

Since the \(\displaystyle x-3\) term can be cancelled, there is a removable discontinuity, or a hole, at \(\displaystyle x=3\).

The remaining denominator of \(\displaystyle x+3\) indicates a vertical asymptote at \(\displaystyle x=-3\).

By looking at the denominator of \(\displaystyle y= \frac{6x-9}{2x-3}\), there will be a discontinuity.

Since the denominator cannot be zero, set the denominator not equal to zero and solve the value of \(\displaystyle x\).

\(\displaystyle 2x-3\neq 0\)

\(\displaystyle 2x\neq 3\)

\(\displaystyle x\neq\frac{3}{2}\)

There is a discontinuity at \(\displaystyle x=\frac{3}{2}\).

To determine what type of discontinuity, check if there is a common factor in the numerator and denominator of \(\displaystyle y= \frac{6x-9}{2x-3}\).

Since the common factor is existent, reduce the function.

\(\displaystyle y= \frac{6x-9}{2x-3}=\frac{(3)(2x-3)}{2x-3}=3\)

Since the \(\displaystyle 2x-3\) term can be cancelled, there is a removable discontinuity, or a hole, at \(\displaystyle x=\frac{3}{2}\).

Start by factoring the numerator and denominator of the function.

\(\displaystyle m(x)=\frac{x^2+4x-5}{x^2+7x+10}=\frac{(x+5)(x-1)}{(x+5)(x+2)}=\frac{x-1}{x+2}\)

A point of discontinuity occurs when a number \(\displaystyle a\) is both a zero of the numerator and denominator.

Since \(\displaystyle x=-5\) is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the \(\displaystyle y\) value, plug in \(\displaystyle -5\) into the final simplified equation.

\(\displaystyle \frac{-5-1}{-5+2}=2\)

\(\displaystyle (-5, 2)\) is the point of discontinuity.

Start by factoring the numerator and denominator of the function.

\(\displaystyle p(x)=\frac{x^3-4x^2-5x}{x^2-9x+20}=\frac{x(x^2-4x-5)}{(x-5)(x-4)}=\frac{x(x-5)(x+1)}{(x-5)(x-4)}=\frac{x(x+1)}{x-4}\)

A point of discontinuity occurs when a number \(\displaystyle a\) is both a zero of the numerator and denominator.

Since \(\displaystyle x=5\) is a zero for both the numerator and denominator, there is a point of discontinuity there. To find the \(\displaystyle y\) value, plug in \(\displaystyle 5\) into the final simplified equation.

\(\displaystyle \frac{5(5+1)}{5-4}=30\)

\(\displaystyle (5, 30)\) is the point of discontinuity.

Start by factoring the numerator and denominator of the function.

\(\displaystyle r(x)=\frac{16-4x^2}{x^2-4}=\frac{-4(x^2-4)}{(x+2)(x-2)}=\frac{-4(x-2)(x+2)}{(x-2)(x+2)}=-4\)

A point of discontinuity occurs when a number \(\displaystyle a\) is both a zero of the numerator and denominator.

Since \(\displaystyle x=\pm2\) is a zero for both the numerator and denominator, there is a point of discontinuity there. Since the final function is \(\displaystyle r(x)=-4\), \(\displaystyle (2, -4)\) and \(\displaystyle (-2, -4)\) are points of discontinuity.