Synthetic division is a short cut for doing long division of polynomials and it can only be used when divifing by divisors of the form . The result or quoitient of such a division will either divide evenly or have a remainder. If there is no remainder, then the "" is said to be a factor of the polynomial. The polynomial must be in standard form (descending degree) and if a degree is skipped such as  it must be accounted for by a "place holder". 

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where  is the remainder.  

While doing the long division we add vertically and we multiply diagonally by k. The empty lines represent places we put the sums and products. Notice that after the first term in the top row there is a 0; this is the place holder. This is because the degrees in the polynomial skipped. When the new coefficients have been found always rewrite starting with one order lower than the highest degree of the original polynomial.

Use synthetic division to verify each factor of the form . Lets start with .

Two goes into 6 three times resulting in:

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From here we see  will give you a remainder of zero and is therefore a factor of the polynomial .

Recall that dividing a polynomial by  does not always result in a pefect division (remainder of 0). Sometimes there is a remainder just like in normal division. When there is a remainder, we write the answer in a certain way.

For example 

 where the divisor is , the quotient or answer is , the remainder is , and the dividend is .

Even though we have variables here, this is the same as noting that  with a remainder of .

And how do we check to know if we have the right answer? We multiply  and add 3 to get 15, our dividend. The same method is used for synthetic division.

Thus, for our problem:

 ,

we must first multiply the divisor by the quotient using the foil method (first multiplying everything in the divisor by x and then everything in the divisor by 3) 

= 

now we just add the remainder which is 1 to yield  which matches the original dividend and is therefore our answer!

To determine if  is a root of the function given, you can use synthetic division to see if it goes in evenly. To set up the division problem, set up the coefficients of the function and then set 1 outside. Bring down the 1 (of the coefficients. Then multiply that by the  being divided in. Combine the result of that  with the next coefficient , which is . Then, multiply that by . Combine that result  with the next coefficient , which gives you . Multiply that by , which gives you . Combine that with the last coefficient , whcih gives you . Since this is not , you have a remainder, which means that  does not go in evenly to this function and is not a root.

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

We multiply what's below the line by  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.

 with remainder 

This can be rewritten as

Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.

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

We multiply what's below the line by  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.

 with remainder 

This can be rewritten as:

Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.

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

We mulitply what's below the line by 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. 

 with remainder 

Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.

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

Remember to place a  when there isn't a coefficient given.

We multiply what's below the line by  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.

 with remainder 

This can be rewritten as:

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 , namely .  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 .  Similarly, our second column transitions from quadratic to linear, making our 5 become .  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 

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.

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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:

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We can interpret this answer as meaning

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  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.

 with reminder 

This can be rewritten as: