Synthetic division

Synthetic division is a shortcut method of performing polynomial long division.

These notes constitute an overview of the topic.

Introduction
Synthetic division is commonly used when dividing a polynomial (such as $$x^3+8x^2-3x+6$$) by a binomial (such as $$x-2$$). The first step is to set the binomial equal to zero and solve for $$x$$ ($$x-2=0$$, $$x=2$$). The next step is to set up the synthetic division using this solution and the coefficients of the polynomial. In this case, the setup would look like this:



It is important to make sure all the numbers are correctly positive or negative depending on the polynomial. Also, if the exponents of the polynomial do not decrease by one from one term to the next, a coefficient of zero must be included. For example, the setup for $$x^3+8x^2-3x+6$$ would use the coefficients 1, 8, -3, and 6.

The next step is to bring down the first coefficient, like this:



After that, the number at the bottom is multiplied by the solution from the binomial (in this case, $$2*1=2$$):



The two numbers above each other are then added:



Again, the new number at the bottom is multiplied by the solution of the binomial:



The two numbers are added:

Again, multiplication:

And addition:



These numbers are the coefficients of the new polynomial. The highest exponent of this new polynomial is one less than that of the original polynomial, so since the original polynomial started with $$x^3$$, the new one will begin with $$x^2$$. The final term is obtained in a slightly different way. The remaining coefficient (in this case, 40) is divided by the original binomial to obtain the last term. Therefore, the final answer is

$$\frac{x^3+8x^2-3x+6}{x-2}=x^2+10x+17+\frac{40}{x-2}$$

This may look like a long process, but in practice it is much faster and simpler than long polynomial division.

Uses
One major use of synthetic division is finding the roots of a polynomial. Synthetic division is an easy way to do this. In the polynomial from the previous example, $$x=2$$ was not a root of the polynomial, because there was a remainder left over after the division. However, if we, for example, take the polynomial $$x^3+3x^2-3x+4$$ and divide it by the binomial $$x+4$$ (when $$x=-4$$)



Since the remainder of this equation is 0, we have found $$x=-4$$ to be a root of the polynomial. We can then take the remaining quadratic $$x^2-x+1$$ and solve for the remaining two roots using the quadratic formula.