Monthly Archives: January 2018

Synthetic Division with a Quadratic Divisor

In class and in the book, it is said that synthetic division only works if the divisor is a linear function, i.e., something that can be written in the form (x + k). That’s not true, but synthetic division with higher-order polynomials is a little more complicated.

Here’s an example of synthetic division using a linear divisor. Consider \((3x^3 + 2x^2 – x + 7)\div (x – 2)\). Here it is using synthetic division: \[\begin{array}{cccccc}2&|&3&2&-1&7\\&|&\downarrow&6&16&30\\&&——&——&——&——\\&&3&8&15&37\end{array}\]

The quotient is \(3x^2 + 8 + 15 + \frac{37}{x – 2}\).

Synthetic division for higher order divisors relies on the same concept, but needs a new line for each coefficient. For instance, a quadratic divisor uses two lines instead of one; a cubic divisor uses three lines; and so on.

For instance, \((x^2 + 3x – 7)(x^2 – 5x + 2) = x^4 – 2x^3 -20x^2 + 41x – 14\). To get a remainder, we’ll change the last two terms. To the left, we’ll reverse the signs of both of the coefficients of the lower terms of the divisor; as before, we’ll write all of the coefficients of the dividend: \[\begin{array}{cccccccc}-3&7&|&1&-2&-20&40&-10\\ & &|&\downarrow&-3&15&-6& \\ & &|&\downarrow&&7&-35&14\\ & & &——&——&——&——&——\\ & & &1&-5&2&-1&4\end{array}\]
In the second row, we use opposite of the coefficient from the \(x\) term (that is, -3) as a multiplier; in the third row, we use opposite of the constant (that is, 7).

This gives us a result of \(x^2 – 5x + 2 + \frac{-x + 4}{x^2 + 3x – 7}\), which is the same thing we get from long division.

Notice that, regardless, the lead coefficient (\(a\)) of the divisor must be 1. You can use synthetic division with, say, a divisor of \(4x – 5\), but you’d have to use \(x – 5/4\) instead, and then divide each coefficient of the result by 4.

For example, consider \((2x^3 + x^2 – 17x + 14)/(2x + 7)\). This is what synthetic division yields: \[\begin{array}{cccccc}-7/2&|&2&1&-17&14\\&|&\downarrow&-7&21&-14\\&&——&——&——&——\\&&2&-6&4&0\end{array}\]

The actual quotient is \(x^2 – 3x + 2\), with no remainder, which is what we get from dividing each of the resulting values by 2.