Polynomial long division
From Academic Kids

In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of lower degree, a generalized version of the familiar arithmetic technique called long division. It can be done easily by hand, because it separates an otherwise complex division problem into smaller ones.
For any polynomials f(x) and g(x), g(x) being of lesser degree than f(x), there exist unique polynomials q(x) and r(x) such that
 <math>\frac{f(x)}{g(x)}=q(x) + \frac{r(x)}{g(x)}<math>
Synthetic division will find the quotient q(x) and remainder r(x) given a dividend f(x) and divisor g(x). The problem is written down like a regular (nonalgebraic) long division problem:
 <math>g(x)\overline{\vert f(x)}<math>;
All terms with exponents less than the largest one must be written out explicitly, even if their coefficients are zero.
Contents 
1 Synthetic division 4 Extension of synthetic division to higher orders 
Example
Find:
 <math>\frac{x^3  12x^2  42}{x3}<math>
The problem is written like this (note that, as explained above, the x term is included explicitly, regardless of the coefficient):
 <math>x3\overline{\vert x^3  12x^2 + 0x  42}<math>
1. Divide the first term of the dividend by the highest term of the divisor. Place the result above the bar (x^{3} ÷ x = x^{2}).
 <math>
\begin{matrix} x^2\\ \qquad\qquad\quad x3\overline{\vert x^3  12x^2 + 0x  42} \end{matrix} <math>
2. Multiply the divisor by the result you just obtained (the first term of the eventual quotient). Write the result under the first two terms of the dividend (x^{2} * (x3) = x^{3}  3x^{2}).
 <math>
\begin{matrix} x^2\\ \qquad\qquad\quad x3\overline{\vert x^3  12x^2 + 0x  42}\\ \qquad\;\; x^3  3x^2 \end{matrix} <math>
3. Subtract the product you just obtained from the appropriate terms of the original dividend, and write the result underneath. This can be tricky at times, because of the sign. ((x^{3}12x^{2})  (x^{3}3x^{2}) = 12x^{2} + 3x^{2} = 9x^{2}) Then, "pull down" the next term from the dividend.
 <math>
\begin{matrix} x^2\\ \qquad\qquad\quad x3\overline{\vert x^3  12x^2 + 0x  42}\\ \qquad\;\; \underline{x^3  \quad3x^2}\\ \qquad\qquad\qquad\quad\; 9x^2 + 0x \end{matrix} <math>
4. Repeat the last three steps, except this time use the two terms that you have just written as the dividend.
 <math>
\begin{matrix} \; x^2  9x\\ \qquad\quad x3\overline{\vert x^3  12x^2 + 0x  42}\\ \;\; \underline{\;\;x^3  \;\;3x^2}\\ \qquad\qquad\quad\; 9x^2 + 0x\\ \qquad\qquad\quad\; \underline{9x^2 + 27x}\\ \qquad\qquad\qquad\qquad\qquad 27x  42 \end{matrix} <math>
5. Repeat step 4. This time, there is nothing to "pull down".
 <math>
\begin{matrix} \qquad\quad\;\, x^2 \;  9x \quad  27\\ \qquad\quad x3\overline{\vert x^3  12x^2 + 0x  42}\\ \;\; \underline{\;\;x^3  \;\;3x^2}\\ \qquad\qquad\quad\; 9x^2 + 0x\\ \qquad\qquad\quad\; \underline{9x^2 + 27x}\\ \qquad\qquad\qquad\qquad\qquad 27x  42\\ \qquad\qquad\qquad\qquad\qquad \underline{27x + 81}\\ \qquad\qquad\qquad\qquad\qquad\qquad\;\; 123 \end{matrix} <math>
The polynomial above the bar is the quotient, and the number left over (123) is the remainder.
 <math>\frac{x^3  12x^2  42}{x3} = x^2  9x  27  \frac{123}{x3}<math>
This method is entirely reminiscent of the long division learned in elementary arithmetic classes  in fact, that technique is simply a special case of polynomial long division, where x=10.
Synthetic division
Synthetic division is a method of performing polynomial long division without having to maintain long records of the process of long division as above  though the processes are still the same. It however, only deals with division by monic linear polynomials, that is, binomials of the form (xb), where b is any rational number.
The change of sign of b when dividing by (x+b) (i.e. b is written to the left of the bar rather than b) allows us to do additions rather than subtractions that are found in regular long division. This reduces the chance for error when the division is done by hand.
Synthetic division is also called division through Ruffini's rule and was described by Paolo Ruffini in 1809.
Performing the same example as before:
 <math>x3\overline{)x^3  12x^2 + 0x  42}<math>
We concern ourselves basically with the coefficients. We write
 <math>\begin{matrix}
3 &  & 1 & 12& 0 & 42 \end{matrix}<math>
Note the change of sign from 3 to 3. Drop the first coefficient after the bar.
 <math>\begin{matrix}
3 &  & 1 & 12& 0 & 42 \\ &  & & & & \\ &  & 1 & & & \\ \end{matrix}<math>
Multiply the dropped number by the number before the bar, and place it in the next column.
 <math>\begin{matrix}
3 &  & 1 & 12& 0 & 42 \\ &  & & 3 & & \\ &  & 1 & & & \\ \end{matrix}<math>
Perform an addition in that column.
 <math>\begin{matrix}
3 &  & 1 & 12& 0 & 42 \\ &  & & 3 & & \\ &  & 1 & 9 & & \\ \end{matrix}<math>
Repeat the previous two steps, the following is obtained
 <math>\begin{matrix}
3 &  & 1 & 12& 0 & 42 \\ &  & & 3 & 27 & 81 \\ &  & 1 & 9 & 27 & 123 \\ \end{matrix}<math>
All the numbers on the last row besides the farthest right correspond to coefficients in the quotient; the last number indicates a remainder. The terms are written with increasing degree from right to left, starting to the left of the remainder with degree 0.
The result of our division is:
 <math>\frac{x^3  12x^2  42}{x  3} = x^2  9x  27  \frac{123}{x  3}<math>
Higher degree synthetic division
The technique of synthetic division described above works only with linear divisors. A similar shortcut method exists for dividing by a quadratic or higher degree monic polynomial.
For instance, let's try to perform the following division:
 <math>\frac{x^3  12x^2  42}{x^2 + x  3}<math>
Create the table as with ordinary synthetic division, using the negative coefficients of the divisor following the leading term. We write
 <math>\begin{matrix}
1 & 3 &  & 1 & 12& 0 & 42 \end{matrix}<math>
Next, underline the leading coefficient of the righthand side, multiply it by the lefthand coefficients and write the products beneath the next columns on the right.
 <math>\begin{matrix}
1 & 3 &  & \underline{1} & 12& 0 & 42 \\ & &  & & 1& 3 & \\ \end{matrix}<math>
Now perform an addition.
 <math>\begin{matrix}
1 & 3 &  & \underline{1} & 12& 0 & 42 \\ & &  & & 1& 3 & \\ & &  & & 13& 3 & 42 \\ \end{matrix}<math>
Repeat the preceding two steps upon the bottom row of the righthand side.
 <math>\begin{matrix}
1 & 3 &  & \underline{1} & 12& 0 & 42 \\ & &  & & 1& 3 & \\ & &  & & \underline{13}& 3 & 42 \\ & &  & & & 13 & 39 \\ & &  & & & 16 & 81 \\ \end{matrix}<math>
The underlined numbers correspond to the coefficients of the quotient, the remaining numbers in the bottom row correspond to the coefficients of the remainder. The terms are written with increasing degree from right to left, with the last terms of the quotient and remainder each having degree 0.
The result of our division is:
 <math>\frac{x^3  12x^2  42}{x^2 + x  3} = x  13 + \frac{16x  81}{x^2 + x  3}<math>
Further explanation for synthetic division
"Synthetic Division" is a simplified, shorthand version of long division of polynomials. There have been suggestions that the terminology originates from the use of "adding negative numbers" instead of "subtracting positive numbers". A series of concrete examples will gradually change to an abstract version, and will show how this method of division is developed. Consider the following examples:
Example 1
Divide (x^2 + 5x + 6) by (x  2), and find its remainder.
One common procedure is to do the following:
x + 7 _______________ x  2  x^2 + 5x + 6 x^2  2x  7x + 6 7x  14  20
Therefore, the remainder of the division is 20.
Notice that we can write this as (x^2 + 5x + 6) = (x  2)(x + 7) + 20. This is written in the form "dividend = (divisor)(quotient) + remainder". Observe also that if f(x) = (x^2 + 5x + 6), then f(2) = 20.
Example 2
To shorthand the division, we may consider removing the x's, and only use the polynomial's coefficients in the division. The same division above can now be written as:
1 7 __________ 1 2  1 5 6 1 2  7 6 7 14  20
Reading off the numbers, we obtain again (x^2 + 5x + 6) = (x  2)(x + 7) + 20.
This division works in the same manner in that we do long division with numbers. Each new row is determined by subtracting the bottom from the top. However, if we consider subtraction as "addition of a negative", then we can simplify the calculation steps a little bit. Compare examples 3 and 4:
Example 3
Divide (x^2  15x + 6) by (x  3). x  12 __________________ x  3  x^2  15x + 6 x^2  3x   12x + 6  12x + 36  30
It is easy to make mistakes when we subtract negative numbers. To avoid this, we negate the divisor, making the intermediate steps addition, instead of subtraction.
Example 4
Divide (x^2  15x + 6) by (x  3). x  12 __________________ x +3  x^2  15x + 6 x^2 + 3x   12x + 6 + 12x  36  30
Notice that the same result is still obtained. Of course, if we write the division in the form "dividend = (divisor)(quotient) + remainder", the divisor must be negated back to (x  3). Rewriting the above division with only coefficients,
Example 5
1 12 __________________ 1 +3  1 15 +6 1 + 3  12 +6 +12 36  30
The column containing 1 and 1, and 12 and +12 are somewhat redundant, since in the division process the goal is to get rid of them. In fact, in synthetic division, we disregard the coefficient of the leading power of x in the divisor, since it is assumed to be 1 (other cases will be taken care of later). Shorthanding our division, this is what is left:
Example 6
1 12 _______________ +3  1 15 +6 + 3  12 +6 36  30
We can collapse our division notation further if the "bringing down the +6" step is eliminated, since it is only a copy of our dividend. In this case, we move the 36 up to replace the +6 so save space
Example 7
1 12 _______________ 1 +3  1 15 +6 + 3  12 36  30
Consider rewriting the above as:
3  1 15 6  3 36  1 12 30
This is the most familiar form of synthetic division. Most of the clutter from the initial long division has been removed. Of course, all the leading 1's can be removed too, but for convenience it is only removed from the dividend.
This has definitely been a leap from concrete to abstract. However, the steps above show that synthetic division does not come "out of thin air". Consider these steps to do synthetic division with the divisor being first degree.
Example 8
1) Write the dividend polynomial in descending powers of x. Any missing powers of x should be filled in with a zero. Then, disregard all the x's and only write out the numbers, in that order.
For example, if you have 1  2x^4 + 1235x^2  3x, then we would write this as "2, 0, 1235, 3, 1", or simply "2 0 1235 3 1". Special care should be taken to write an extra "0" if the factor in front of x to the zeroth power is zero. So, x^3 + 3x should be written as "1 0 3 0".
2) The divisor (for now), must be put into the form x  c. In synthetic division, we will only write "a" to represent the dividend.
For example, (x  3) is written as "3", while (x + 4) is written as "4". Think of it as writing down the value for which (x  c) will equal zero.
3) Write out the division in this format:
divisor  dividend  <blank spaces here> ____________________
So, if we wanted to divide x^4 + 2x^3  3x + 12 by x + 5, we would write
5  1 2 0 3 12  ________________
Remember, the "0" represents the "imaginary" term 0x^2 in the polynomial x^4 + 2x^3  3x + 12.
4) Bring down the first number in the dividend, and write it below the line.
Using our example above,
5  1 2 0 3 12  ________________ 1
5) Multiply this first number by the number representing the divisor. Then, write the result in the next column above the horizontal line.
5  1 2 0 3 12  5 ________________ 1
6) Add the two numbers in the second column. Write the result below th line.
5  1 2 0 3 12  5 ________________ 1 3
7) Multiply this number by the divisor, and write it in the next column above the line.
5  1 2 0 3 12  5 15 ________________ 1 3
8) Add the numbers in the new column together and write the result below the horizontal line.
5  1 2 0 3 12  5 15 ________________ 1 3 15
9) Multiply this number by the divisor, and write it in the next column above the line. (Notice how the steps are repeating)
5  1 2 0 3 12  5 15 75 _________________ 1 3 15
10) Add the numbers in the new column together and write the result below the horizontal line.
5  1 2 0 3 12  5 15 75 _________________ 1 3 15 78
11) Multiply this number by the divisor, and write it in the next column above the line.
5  1 2 0 3 12  5 15 75 390 _________________ 1 3 15 78
12) Add the numbers in this new column together and write the result below the horizontal line.
5  1 2 0 3 12  5 15 75 390 _________________ 1 3 15 78 402
13) There are no more columns to write numbers in. Since our dividend was of degree 4, and divisor of degree 1, it should be expected that the quotient should be of degree 3. Therefore, we save the last number 412, as the remainder of the division. An "L" may be placed to distinguish the remainder from the quotient like so:
5  1 2 0 3 12  5 15 75 390 __________________ 1 3 15 78 402 
14) The process of synthetic division stops here. By inspection then, the result
(x^4 + 2x^3 3x + 12) = (x + 5)(x^3  3x^2 + 15x  78) + 402 is obtained. Verify the result if desired. With practice, this method is very fast at finding remainders of polynomials upon division. Review this example if needed.
Example 9
Here is another shorter example:
Divide (x^2  3x  4) by (x + 1)
Dividend: 1 3 4 Divisor: 1
1  1 3 4  1 4  1 4  0 
Observe that 0 is a remainder. Therefore, we have found through synthetic division that (x + 1) is a factor of (x^2  3x 4). In fact, (x^2  3x 4) = (x + 1)(x  4).
Observe that if we use (x  4) as the divisor in the above example, the same result is obtained.
Example 10
Divide (x^2  3x  4) by (x  4)
Dividend: 1 3 4 Divisor: 4
4  1 3 4  4 4 > (x^2  3x  4) = (x  4)(x + 1), as shown above.  1 1  0 
Example 11
Here is another example for consolidation:
Divide f(x) = (x^5  2x^4 + x^3  2x^2 3x + 1) by (x + 1)
Dividend: 1 2 1 2 3 1 Divisor: 1
1  1 2 1 2 3 1  1 3 4 6 3  1 3 4 6 3 2 
Compare these numbers with the intermediate numbers obtained from Horner's Method for evaluating polynomials:
f(x) = (x^5  2x^4 + x^3  2x^2 3x + 1) = ((((x  2)x + 1)x  2)x  3)x + 1
Now,
f(1) = [[[[(1)  2](1) + 1](1)  2](1)  3](1) + 1 = [[[(3)(1) + 1](1)  2](1)  3](1) + 1 = [[(4)(1)  2](1)  3](1) + 1 = [(6)(1)  3](1) + 1 = (3)(1) + 1 = 2
Synthetic division can be considered to be a condensed version of Horner's Method for evaluation of polynomials.
So far, the technique seems to be promising, but very limited because we can only divide by first degree polynomials of degree 1 of the form (x  c). However, it is possible to extend the idea to division by polynomials of nth degree, as long as the dividend has a higher power than its divisor.
First, consider the two theorems:
Theorem 1: Remainder Theorem Let f(x) be a polynomial in x. Then, the remainder when f(x) is divided by (ax  b) is f(b/a).
Proof of Theorem 1: Let f(x) = g(x)(ax  b) + R, where R is a degree zero remainder, and g(x) is the quotient. Then, f(b/a) = g(b/a)(0) + R, or, after simplification, f(b/a) = R. This is the result stated above.
Theorem 2: Factor Theorem Let f(x) be a polynomial in x. If f(b/a) = 0, then (ax  b) is a factor of f(x). Conversely, if (ax  b) is a factor of f(x), then f(a/b) = 0.
Proof of Theorem 2: From above, it has been shown that f(x) = g(x)(ax  b) + f(b/a). If f(b/a) = 0, then f(x) = g(x)(ax  b). Therefore (ax  b) is a factor of f(x). The converse also follows a similar proof. From above, we defined f(x) = g(x)(ax  b) + f(a/b). If (ax  b) is a factor of f(x), then f(a/b) must equal zero. Otherwise, we would have a remainder, in which case (ax  b) would not be a factor of f(x). This completes the proof.
Consider first degree polynomials of the form (ax  b). If (ax  b) is a factor of f(x), then (x  b/a) must also be a factor, since f(b/a) = 0 if and only if (ax  b) is a factor of f(x). From this result, we may use fractions to indicate division by such (ax  b) factors.
Example 12
Example: Divide (4x^3 + 4x^2  57x + 30) by (2x  1).
Since (2x  1) is not of the form (x  c), we will use (x  1/2) as the divisor, since 2(x  1/2) = (2x  1). Therefore, our divisor in synthetic division is 1/2.
Dividend: 4 4 57 30
1/2  4 4 57 30  2 3 27  4 6 54  3 
Now, the complication here is that since 1/2 represents (x  1/2), which is only half of the factor (2x  1) the division makes results in a quotient that is twice the actual result. We have to divide this by two then. However, we disregard the quotient.
So,
1/2  4 4 57 30  2 3 27  4 6 54  3 
becomes
1/2  4 4 57 30  2 3 27  4 6 54  3  2 3 27 Therefore, (2x  1)(2x^2 + 3x  27) + 3 = (4x^3 + 4x^2  57x + 30)
Example 13
Another example: Divide (9x^2 + 39x + 42) by (3x + 7)
Divisor: Since 3(x + 7/3) = (3x + 7), in synthetic division, (3x + 7) is written as 7/3.
Dividend: 9 39 42
7/3  9 39 42  21 42  9 18  0 
But remember, 7/3 represents (x + 7/3), which is a third of (3x + 7), so we divide our quotient by 3.
7/3  9 39 42  21 42  9 18  0  3 6  0 
Notice that in the line 3 6 0, the 3 and 6 have a greatest common factor greater than 1, namely 3. This indicates that the original polynomial (9x^2 + 39x + 42) had a factor of 3 to begin with. Whenever possible, do not overlook these simple steps that can reduce the complexity of your calculations.
So,
7/3  9 39 42  21 42  9 18  0  3 6  0 
becomes
7/3  9 39 42  21 42  9 18  0  3 6  0 / 3 +   1 2  0     Immediately, (9x^2 + 39x + 42) = 3(x + 2)(3x + 7)
Example 14
For consolidation:
Divide [(1/3)x^2 + (1/6)x + (1/2)] by (4x + 1)
Divisor:  1/4 Dividend: 1/3 1/6 1/2 1/4  1/3 1/6 1/2  1/12 1/48  1/3 1/12  23/48 
However, remember that 1/4 is actually 1/4th of our divisor. So, the division becomes:
1/4  1/3 1/6 1/2  1/12 1/48  1/3 1/12  23/48  1/12 1/48  23/48 
Indeed, (4x + 1)[(1/12)x + (1/48)] + 23/48 = (1/3)x^2 + (1/6)x + 1/2.
Extension of synthetic division to higher orders
Extension of synthetic division into divisors of higher degrees is also not too difficult.
1) Let the divisor be a polynomial of the form:
f(x) = (a_n)x^n + (a_n1)x^(n1) + (a_n2)x^(n2) + ... + (a_2)x^2 + (a_1)x + a_0
and rewrite this polynomial in "synthetic division form". So, f(x) becomes
(a_n) (a_n1) (a_n2) ... (a_2) (a_1) (a_0)
For example, (2x^2  4x + 3) becomes (2 4 3).
2) Synthetic division requires the leading coefficient be 1, so divide each number by (a_n), and get
1 (b_n1) (b_n2) ... (b_2) (b_1) (b_0)
where each (b_ni) = (a_ni) / (a_n), where 1 <= i <= n. Remember this step as this will be referred to later on.
Using the example, (2 4 3) now becomes (1 2 3/2).
3) Drop the leading 1 as the coefficent and negate all the terms in its synthetic division representation. So,
1 (b_n1) (b_n2) ... (b_2) (b_1) (b_0)
becomes
(b_n1) (b_n2) ... (b_2) (b_1) (b_0)
Referring to the example, (1 2 3/2) becomes (2 3/2).
4) Write the divisor along the vertical bar in synthetic division such that the coefficients representing the divisor are in descending powers of x going from top to bottom.
So, the divisor (2 3/2) is written as:
2  not 3/2  3/2  2   
5) Suppose that the dividend is 2x^3  8x^2 + 13x  9. Then in synthetic division form, the division is written as
 2 8 13 9 2  3/2  
This is in the form
divisor  dividend  <blank spaces here>  <another blank space here> ___________________________
There are two blank spaces because two numbers are used to represent the divisor. If 5 numbers were used, then there would be 5 blank rows.
6) Bring down the first number in the dividend and write it below the horizontal line.
 2 8 13 9 2  3/2   2
7) Call this number "a". Multiply "a" with each of the numbers in the divisor in order, going from top to bottom, and write the results of the multiplications diagonally going from top to bottom, with the first result being written in the second column
2 times 2 is 4. 2 times 3/2 is 3.
 2 8 13 9 2  4 3/2  3  2
8) Add all the numbers in the second row, and write the result below the horizontal line.
 2 8 13 9 2  4 3/2  3  2 4
9) Steps 7 and 8 are the repeated steps. So, using "4" as our "pivot",
4 times 2 is 8. 4 times 3/2 is 6.
Write these numbers in diagonally, going from top to bottom, starting in the third row.
 2 8 13 9 2  4 8 3/2  3 6  2 4
10) Add all the numbers in the third row, and write the result below the horizontal line.
 2 8 13 9 2  4 8 3/2  3 6  2 4 2
11) Since our division process has taken us to the last column, stop. Add all the remaining numbers in the columns that have not been added. In this case, it is just the last column containing 9 and 6. Add them together and write the result below the horizontal line.
 2 8 13 9 2  4 8 3/2  3 6  2 4 2 3
12) Since we divided by a divisor with two numbers, the last two numbers in the division are saved as the remainder. The reason is that the degree of the dividend was 3, and the degree of the divisor was 2, which means the degree of the remainder must be 3  2 = 1. In synthetic division, such a polynomial is represented by 2 numbers, so the last two are saved in the remainder.
 2 8 13 9 2  4 8 3/2  3 6  2 4  2 3 < 2 3 is remainder 
14) Since 2 3/2 represents (x^2  2x + 3/2) which is only half of (2x^2  4x + 3), we need to divide our quotient (2 4) by 2.
 2 8 13 9 2  4 8 3/2  3 6  2 4  2 3  1 2  2 3 
15) From above, the result
(2x^3  8x^2 + 13x  9) = (2x^2  4x + 3)(x  2) + 2x  3
is obtained.
Here is a longer example for consolidation:
Divide (3x^7 + 8x^6  10x^5 + 4x^4 + 2x^3  9x^2 + 6x + 5) by (2x^3 + 4x^2  5x + 1/2)
Divisor: (2x^3 + 4x^2  5x + 1/2) > (2 4 5 1/2) > (1 2 5/2 1/4) > (2 5/2 1/4)
so (2 5/2 1/4) is our divisor. Remember that this is 1/2 of our actual divisor, so the final quotient needs to be multiplied by 1/2 when the division process is finished.
Dividend: 3 8 10 4 2 9 6 5
1) Write into the synthetic division format
 3 8 10 4 2 9 6 5 2  5/2  1/4  
2) Bring down the first number in the dividend.
 3 8 10 4 2 9 6 5 2  5/2  1/4   3
3) Multiply this number (3) by all the coefficients in the divisor.
_ 3 times 2 is 6  3 times 5/2 is 15/2 > Write these numbers in like so: 3 times 1/4 is 3/4 _  3 8 10 4 2 9 6 5 2  6 5/2  15/2 1/4  3/4  3
4) Add all the numbers in the second column and write the result below the horizontal line.
 3 8 10 4 2 9 6 5 2  6 5/2  15/2 1/4  3/4  3 2
4) Multiply this number (2) by all the coefficients in the divisor.
_ 2 times 2 is 4  2 times 5/2 is 5 > Write these numbers in like so: 2 times 1/4 is 1/2 _
 3 8 10 4 2 9 6 5 2  6 4 5/2  15/2 5 1/4  3/4 1/2  3 2
5) Add all the numbers up in the third column and write the result below the horizontal line.
 3 8 10 4 2 9 6 5 2  6 4 5/2  15/2 5 1/4  3/4 1/2  3 2 13/2
6) Multiply this number (13/2) by all the coefficients in the divisor.
_ 13/2 times 2 is 13  13/2 times 5/2 is 65/4 > Write these numbers in like so: 13/2 times 1/4 is 13/8 _
 3 8 10 4 2 9 6 5 2  6 4 13 5/2  15/2 5 65/4 1/4  3/4 1/2 13/8  3 2 13/2
7) Add all the numbers up in the fourth column and write the result below the horizontal line.
 3 8 10 4 2 9 6 5 2  6 4 13 5/2  15/2 5 65/4 1/4  3/4 1/2 13/8  3 2 13/2 85/4
8) Multiply this number (85/4) by all the coefficients in the divisor.
_ 85/4 times 2 is 85/2  85/4 times 5/2 is 425/8 > Write these numbers in like so: 85/4 times 1/4 is 85/16 _
 3 8 10 4 2 9 6 5 2  6 4 13 85/2 5/2  15/2 5 65/4 425/8 1/4  3/4 1/2 13/8 85/16  3 2 13/2 85/4
9) Add all the numbers in th 5th column and write the result below the horizontal line
 3 8 10 4 2 9 6 5 2  6 4 13 85/2 5/2  15/2 5 65/4 425/8 1/4  3/4 1/2 13/8 85/16  3 2 13/2 85/4 275/4
10) Multiply this number (275/4) by all the coefficients in the divisor.
_ 275/4 times 2 is 275/2  275/4 times 5/2 is 1375/8 > Write these numbers in like so: 275/4 times 1/4 is 275/16 _  3 8 10 4 2 9 6 5 2  6 4 13 85/2 275/2 5/2  15/2 5 65/4 425/8 1375/8 1/4  3/4 1/2 13/8 85/16 275/16  3 2 13/2 85/4 275/4
11) Since the last number reaches the end of the calculation, add all the numbers each of the last 3
columns together.
 3 8 10 4 2 9 6 5 2  6 4 13 85/2 275/2 5/2  15/2 5 65/4 425/8 1375/8 1/4  3/4 1/2 13/8 85/16 275/16  3 2 13/2 85/4 275/4  733/4 2739/16 355/16 
Now, recalling that there is a factor of 1/2 in the quotient, NOT the remainder. See step 1):
12) Indeed, from the above result,
3x^7 + 8x^6  10x^5 + 4x^4 + 2x^3  9x^2 + 6x + 5  2x^3 + 4x^2  5x + 1/2 = 3/2 x^4 + x^3  13/4 x^2 + 85/8 x  275/8 733/4 x^2 2739/16 x + 355/16 +  2x^3 + 4x^2  5x + 1/2