Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor -- and it only works in this case. Synthetic division is generally used, however, not for dividing out factors but for finding zeroes (or roots) of polynomials. More about this later.
If you are given, say, the polynomial equation y = x2 + 5x + 6, you can factor the polynomial as y = (x + 3)(x + 2). Then you can find the zeroes of y by setting each factor equal to zero and solving. You will find that x = –2 and x = –3 are the two zeroes of y.
You can, however, also work backwards from the zeroes to find the originating polynomial. For instance, if you are given that x = –2 and x = –3 are the zeroes of a quadratic, then you know that x + 2 = 0, so x + 2 is a factor, and x + 3 = 0, so x + 3 is a factor. Therefore, you know that the quadratic must be of the form y = a(x + 3)(x + 2).
(The extra number "a" in that last sentence is in there because, when you are working backwards from the zeroes, you don't know toward which quadratic you're working. For any non-zero value of "a", your quadratic will still have the same zeroes. But the issue of the value of "a" is just a technical consideration; as long as you see the relationship between the zeroes and the factors, that's all you really need to know for this lesson.)
Anyway, the above is a long-winded way of saying that, if x – n is a factor, then x = n is a zero, and if x = n is a zero, then x – n is a factor. And this is the fact you use when you do synthetic division.
Let's look again at the quadratic from above: y = x2 + 5x + 6. From the Rational Roots Test, you know that ± 1, 2, 3, and 6 are possible zeroes of the quadratic. (And, from the factoring above, you know that the zeroes are, in fact, –3 and –2.) How would you use synthetic division to check the potential zeroes? Well, think about how long polynomial divison works. If we guess that x = 1 is a zero, then this means that x – 1 is a factor of the quadratic. And if it's a factor, then it will divide out evenly; that is, if we divide x2 + 5x + 6 by x – 1, we would get a zero remainder. Let's check:
As expected (since we know that x – 1 is not a factor), we got a non-zero remainder. What does this look like in synthetic division? Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved
First, write the coefficients ONLY inside an upside-down division symbol:
