In Part 1, we derived the structure of the procedure for doing Synthetic Division in the Restricted domain used by textbooks dealing with this topic (RSD), and I suggested that the reader might find it useful to write out a formal step-by-step procedure based on the example we used in this derivation. In the current post, I shall begin with such a procedure, then briefly describe a variation acceptable to the standard texts, then give a second procedure that generalizes SD to division by binomials of any degree whose second term is a constant and give three examples of this type of SD in order to clarify possible sources of ambiguity.
The final array will look like this:
It will be generated by following these steps:
- Write the top row, the horizontal bar, and the leftmost entry in the bottom row, which is the same number as the one directly above it in the top row (the leading coefficient of the dividend).
- Write the product of k and the latest entry of the bottom row in the next column of the middle row.
- Write the sum of that product and the coefficient directly above it in the bottom row of the same column.
- Repeat steps 2 and 3 to obtain the next product and sum, and the next, etc., until the sum is written in the last column.
- Reading the result: The degree of the quotient is n – 1, the coefficients of the quotient are the entries in the bottom row except for the last, and the remainder is the last entry of the bottom row, so the exact quotient is:
After a few examples of this method, the textbooks sometimes introduce one variation. To divide a polynomial by mx – k, first divide the divisor by m, yielding x – (k/m), which has the same form as the divisor in RSD, so the division can be performed using the procedure that we already have. To get the correct quotient, however, we must either divide the dividend by m, too, or use the same dividend and divide the resultant quotient by m. For instance: (3x³ + 4x² – 6x + 5) / (2x – 6) = (1.5x³ + 2x² – 3x + 2.5) / (x – 3) or [(3x³ + 4x² – 6x + 5) / (x – 3)] / 2.
The final array will look like this:
As long as m ≤ n, m and n are independent of each other, which leads to notational paradoxes, e.g., if m is more than half of n, then m > n – m and the third and fourth columns in this array are incorrectly ordered. More thought than I’m willing to give to it would be required to eliminate all such notational paradoxes. Fortunately, learning the procedure does not depend on having absolutely correct notation, and following the steps listed below will easily generate the correct array to yield the correct quotient.
- Write the top row, the horizontal bar, and the m leftmost entries in the bottom row, which are the same numbers as the ones directly above them in the top row (the first m coefficients of the dividend).
- Write the products of k and the latest m entries of the bottom row in the next m columns of the middle row.
- Write the sums of those products and the coefficients directly above them in the bottom row of their respective columns.
- Repeat steps 2 and 3 to obtain the next m products and sums, and the next, etc., until the product and sum are written in the last column.
- Reading the result: The degree of the quotient is n – m, the coefficients of the quotient are the first n – m + 1 entries in the bottom row, and the coefficients of the remainder are the last m entries of the bottom row, so the exact quotient is:
This procedure can be derived from the corresponding long division just as the RSD was, and I leave it as an exercise for the reader. Here are some examples of this SD procedure. To see how the array of numbers is produced, write it yourself using the step-by-step algorithmic procedure just outlined.
- Divide (x5 – 2x4 – 6x3 + 10x2 + 11x – 8) by (x2 – 3), the specific divisor that the precalculus textbook that I quoted said could not be used as a divisor in SD.
The quotient is (x3 – 2x2 – 3x + 4) + (2x + 4)/(x2 – 3) or (x3 – 2x2 – 3x + 4) with a remainder of (2x + 4).
2. Divide (3x6 – 4x5 + x4 + 22x3 – 12x2 + 19x + 37) by (x3 + 4).
The quotient is (3x3 – 4x2 + x + 10) + (4x2 + 15x – 3)/(x3 + 4) or (3x3 – 4x2 + x + 10) with a remainder of (4x2 + 15x – 3).
Note: In multiplying the second group of three sums (+10, +4, +15) by -4, there is room in the middle row for only the first product (-40), so we stop there, as instructed to in procedure step 4.
3. Divide (x8 – x7 – x6 + 3x4 + 2x3 – 5) by (x4 – 2).
The quotient is (x4 – x3 – x2 + 5) + (-2x2 + 5)/(x4 – 2) or (x4 – x3 – x2 + 5) with a remainder of (-2x2 + 5).
Note: Since the dividend contains no terms with an exponent of 5, 2 or 1, we insert zeros for the coefficients of those terms. Also, since there are zeros among the quotient and remainder coefficients, we omit the appropriate terms in both.
It’s easy enough to make up exercises like these and it’s easy enough to check your answers by doing the requisite multiplication of quotient by divisor and adding the remainder to the product. As always, questions and comments are welcome.
Next time we’ll post the USD procedure for the general case of dividing one polynomial by another of equal or lesser degree.