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Synthetic division, which was invented in 1809, is a spatial procedure for dividing a polynomial by a binomial of the form x-a. The purpose of this article is to provide and explain the underlying algorithm used in the synthetic division procedure. Nemeth braille is used to write the formulas for the algorithm and for other math expressions. The formulas for the algorithm are given here at the start of the article for the benefit of those with the necessary background and are explained later. If you have any questions you can contact me by email using the address at the bottom of this page.
This is the formula for the division of a polynomial by the binomial x-a. The formula is written in Nemeth braille using the standard ASCII braille encoding.
Equation 1
The c;i are the coefficients in the given polynomial which has n terms. The q;j are the coefficents in the quotient polynomial and r is the remainder. The formulas for calculating the latter items are given in Equations 2-4 below.
The left hand side of the formula is the division to be carried out and the right hand side is the calculated result of the division. The sums in both expressions use a standard general form for representing a polynomial as explained in Section 4. The polynomials are written with the exponents decreasing from left to right and with any missing terms indicated with explicit zeros as detailed in Section 3.
These are the two formulas for calculating the coefficients q;j and the formula for calculating the remainder r needed in Equation 1.
Equations 2-4
q1 .k c1
q;j .k q;j-1"a+c;j for all j ." 1 and ". n
r .k q;n-1"a+c;n
Here's an example of applying the formulas with the three-term polynomial x^2+0-4 as the dividend and x-2 as the divisor. The givens are thus
n .k #3, c1 .k #1, c2 .k #0, c3 .k -#4, and a .k #2
Substituting the givens into Eqs. 2 and 3 for the coefficients of the quotient and into Eq. 4 for the remainder yields the following
q1 .k c1 .k #1
q2 .k q1*a+c2 .k #1*2+0 .k #2
r .k q2*a+c3 .k #2*2-4 .k #0
The result of Eq. 1 is thus q1*x+q2 .k x+2 since the remainder is zero. You can check this result by multiplying it by the divisor and ensuring that it yields the original dividend.
Synthetic division uses a spatial layout intended to make it easy to organize and use all the information needed to carry out the division of polynomials by binomials of the form x-a. (Synthetic division itself isn't described here as it is easy to find descriptions of it elsewhere.) Synthetic division is analogous to the standard spatial procedure for long division of numbers.
Synthetic division can present difficulties for braille users since spatial layouts can be difficult to read and understand in braille. This presents a dilemma for math teachers who are teaching common procedures that are important to further understanding. Moreover, these teachers have likely learned that teaching rote procedures is easier than teaching understanding.
Synthetic division is very similar to long division of numbers. An article by two college professors titled The Role of Long Division in the K-12 Curriculum explains the importance of understanding why and how long division works. This quote from the article outlines some reasons for the authors' claim.
There are at least two separate areas where the insights gained from understanding the long division algorithm are crucial, first calculus, and second, applications of polynomials in advanced areas of mathematics and related fields. ... There is reason to believe that students who wish to enter technical areas but have had only minimal experience with polynomials and none with partial fraction techniques are at a severe disadvantage, and at least some are forced to give up their aspirations.Note that the Appendix to the referenced article has a good explanation of long division that doesn't rely on the common spatial procedure.
The present article gives the formulas for the algorithm used in synthetic division and leaves it to individual students to choose how they prefer to organize the needed information and to apply the formulas. Some have suggested the use of an Excel spreadsheet for this purpose. Another alternative is to design a data structure and implement the formulas in a short program written in your favorite programming language.
It is useful to review the general terminoloy of division before discussing synthetic division. The quantity to be divided is called the dividend. The quantity by which the dividend is to be divided is called the divisor or factor. The result of or answer to a division process has two parts: the quotient and the remainder. (Note that some writers use the term quotient to refer to the entire result but that isn't done here.) You can always use a two-step checking process to find whether a division has been done correctly. First multiply the quotient by the factor to get their product. Second add the product from the first step to the remainder to get the result. If the division process is correct then the result of the checking process will equal the dividend.
Here is the formula for the checking process.
d .k f*q+r
In this formula the result dividend is represented by the symbol letter d, the factor by the symbol letter f, the quotient by the symbol letter q, and the remainder by the symbol letter r.
As an example of simple division using integers let the dividend be #26 and the factor #6. You should be able to do this division in your head. The quotient is #4 and the remainder is #2. Now apply the checking process. The factor #6 times the quotient #4 equals #24 and #24 plus the remainder #2 equals #26. The result dividend equals the original dividend so the division is correct.
In synthetic division the dividend is a polynomial. (It is assumed here that you know what a polynomial is. If not, you may need to first study some background information.) This section explains how to write a polynomial in the special form needed for synthetic division.
Remember that each term of a polynomial consists of a coefficient times a quantity with an exponent. Here we are only concerned with polynomials where the coefficients are positive integers, negative integers, or zero. Sometimes if the coefficient is one it is not written explicitly but when writing a polynomial in the special form needed here all the coefficients including a coefficient of one are written explicitly. Also here the quantity will always be represented by the letter x and the exponent will always be a positive integer. An exponent of one does not need to be written explicitly and a quantity with a zero exponent does not need to be written explicitly since x^0 equals one.
The first step is to collect the terms of the polynomial if necessary. This means to add any terms where the quantities have the same exponent so there is only one term with each exponent.
Example of collecting terms
#4x^3+2x^2+3x+3x^2 .k #4x^3+3x+5x^2
The second step is to rearrange the terms of the polynomial as necessary to ensure that the exponents are decreasing from left to right. If there is a term that is just an integer it must be written last. This is because this term is equivalent to treating the integer as the coefficient of the quantity x^0.
Examples of ordering terms according to exponents
x^3+3x+5x^2 .k x^3+5x^2+3x
x^2-3+x .k x^2+x-3
The largest exponent in a polynomial is called the order or degree of the polynomial. Once the polynomial is rewritten according to the first two steps, there will be at most one more term than the order or value of the largest exponent. If the number of terms is equal to the order plus one this means there are no missing terms. If the number of terms is less than the order plus one this means there are one or more missing terms.
The third step is only necessary if there are any missing terms. In this case it is necessary to insert zeros as placeholders in each position where one or more terms is missing. For example if there is a term with the quantity x^3 and one with the quantity x but not one with the quantity x^2, there needs to be a zero placeholder inserted between the terms in x^3 and x. Also if the last term is not an integer, you will need to add a zero.
Examples of inserting placeholders
x^3+5x^2+3x .k x^3+5x^2+3x+0
x^3+3x .k x^3+0+3x+0
x^3+2 .k x^3+0+0+2
Since the goal here is to present an algorithmic equivalent to synthetic division that doesn't use a spatial layout, we will need to use a general notation that can represent any polynomial which has been written according to the steps in Section 3. The standard way to do this is to use sum notation. The following is intended to be clear even if you've never used sum notation before. However if you have questions, you can send them to the address at the bottom of the page.
One item that our sum notation requires is a general expression that can be used to represent any term of a polynomial. This general expression needs to represent the coefficient and also the exponent of the quantity x.
A standard way to write the coefficient in sum notation is c subscript i where i is the position of the term in the polynomial. (The values of c and the possible values of i will, of course, be different for different polynomials.) The position of a term is also commonly called the index. The leftmost position or first term is position one. s
The general expression for the exponent is based on a rule that you may have to think about for a minute. The rule is that if a suitable polynomial has been properly written according to the steps in Section 3 above, the exponent for each quantity is equal to the total number of terms minus the position of that term. If we use the symbol n for the total number of terms then the rule means that the formula for the exponent for quantity term is n-i where is is the symbol for the position.
Let's illustrate this rule with a simple example of a properly written polynomial, x^2+3x+4, which has three terms. The exponent of x in the first or position one term is #2 which is three minus one. The exponent in the second or position two term is #1 (implied) which is three minus two. The exponent in the third or position three term is #0 which is three minus three. (Since x raised to the zero power equals one and four times one equals four, the factor one hasn't been shown explicitly in the polynomial.)
Multiplying the general expression for a coefficient by the quantity x with the corresponding general expression for its exponent yields this general expression that can be used for any term of the polynomial.
c;i"x^n-i
where i is the position of the term and n is the constant total number of terms in the polynomial.
The other item that our sum notation needs besides a general expression for the terms of a polynomial are the lower and upper limits of the values of the index or position i. For polynomials written as specified in Section 3 the starting value or lower limit of i is always #1 and the ending value is always the number of terms.
Sum notation is written by preceding the general term with a capital Greek sigma and including the lower and upper limits. The actual terms to be added together to represent the polynomial are determined by substituting each value of i from the lower limit to the upper limit into the general expression for a term.
In print the lower limit is typically written in small characters below the sigma and the upper limit is written in small characters above the sigma. In Nemeth braille the limits are handled as modifiers and are written using the Five-Step Rule for Modified Expressions. (This Rule is used here for this case. If you want more information about the Five-Step Rule, you will need to consult a Nemeth manual.)
Our general polynomial is thus written as
".,s<i .k #1%n]c;i"x^n-i
where n is the total number of terms in the polynomial including any placeholder terms where c;i equals zero.
The formulas for the algorithmic equivalent were given at the beginning of this article.
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First posted May 14, 2014.