Synthetic Division is a method of dividing one polynomial by another of equal or lesser degree. If you already know how to do SD, you probably think I am leaving a required restriction out of my definition. That is because all textbooks that deal with SD include that restriction, either by making a fairly innocuous statement such as, “In many polynomial division problems the divisor is a first-degree binomial of the form x – k, where the coefficient of x is 1″, or by making a statement that I will herein show is patently false, such as, “Synthetic division works only for divisors of the form x – k. You cannot use synthetic division to divide a polynomial by a quadratic such as x² – 3.” (These are direct quotes from two different textbooks, the first one on college algebra and the second one on precalculus. And perhaps I should point out that the second quote is false only because the textbook considers the version of SD also known as Horner’s Algorithm, which has been around for 200 years, and not the extended version that I am introducing here.)
We will introduce Universal SD in three stages. First, we will present one version of the derivation of the procedure for the restricted version of SD, so we can see why it works. Then we will extend the procedure to the simple variation mentioned in the second quote. Finally, we will extend it to the general case.
The algorithm in numerical long division is: divide (approximately), multiply, subtract, bring down, and it looks like this: _429 15 ) 6437 Divide 64 by 15 (& write 4 on top) 60 Multiply 15 by 4 (& write 60 under 64) 43 Subtract 60 from 64 (& write 4 under 60) 30 Bring down the 3 (& write it next to the 4) 137 Repeat until you run out of digits in the 135 dividend. 2
The same thing is done in polynomial long division and it looks like this: 2x² + 6x + 7 x – 5 ) 2x³ – 4x² – 23x – 16 . 2x³ – 10x² 6x² – 23x 6x² – 30x 7x – 16 7x – 35 19 Divide 2x³ by x, multiply x – 5 by 2x², subtract 2x³ – 10x² from 2x³ – 4x², bring down -23x. Repeat these four steps until you run out of terms in the dividend. If any term of degree less than the degree of the dividend is not in the dividend, we write it with coefficient 0 in the dividend.
In SD with the same divisor and dividend, to find a shortcut, we eliminate writing anything that isn’t needed to know what is being omitted. So we start by omitting variables with their exponents (leaving only the coefficients of those variables), and the unwritten coefficient 1 of the divisor (since we’re doing only the restricted version of SD first), resulting in this: 2 + 6 + 7 -5 ) 2 – 4 – 23 – 16 . 2 – 10 . 6 – 23 . 6 – 30 7 – 16 7 – 35 19
Each of the top row (quotient) coefficients is repeated twice in the same column and each coefficient that was brought down from the dividend is repeated once. By eliminating all the repetitions except one and writing the top row as the bottom row, we can compress this array of coefficients into this: -5 ) 2 – 4 -23 -16 . -10 -30 -35 . 2 +6 +7 +19
The bottom row can now be seen to be the result of subtracting the middle row from the (new) top row (coefficients of the dividend) and each of the middle row coefficients is the result of multiplying the previous column’s bottom row coefficient by the divisor (-5). If we change the sign of the divisor, thereby changing the signs of the middle row coefficients, we then add instead of subtract, and we get this: 5 ) 2 – 4 -23 -16 +10 +30 +35 . 2 +6 +7 (+19
This array can be derived directly from the divisor and dividend by writing x – 5 as 5 and the coefficients of the dividend in the top row and the first coefficient of the dividend in the bottom row. The rest of the coefficients in the array can be derived by writing the product (+10) of 5 (the “divisor”) and the 2 of the bottom row in the middle row under the -4, writing the sum of the two numbers in that column in the bottom row, and repeating this two-step algorithm until we run out of columns to add. The bottom row contains the coefficients of the quotient and remainder, and since the degree of the quotient is one less than the degree of the dividend, the quotient is 2x² + 6x + 7 with a remainder of 19 and the exact quotient is 2x² + 6x + 7 + 19/(x – 5).
Even if you have never done SD before, from the preceding paragraph you could write your own procedure for this Restricted SD, and try it out on a few problems, which I encourage you to do. In my next post, I’ll start by writing a formal procedure for RSD, briefly present one modification allowed by the current textbooks, and then give the procedure for SD of a polynomial of degree n by a binomial of degree m and whose second term is a constant and where 2 ≤ m ≤ n. In the post after next, I will present my procedure for Universal SD.