Synthetic Division

by Shelley Walsh ©2000

Synthetic division is a shortcut method of doing long division of polynomials when the divisor is of the form x+c. Long division with divisors of this for is the most commonly used kind because it is the long division that is needed to determine whether a number is the solution to an equation. This kind of long division is also useful for substituting in values to a polynomials, because of a theorem called the remainder theorem. The reason synthetic division works is that for this kind of division it is possible to greatly streamline down the process by cutting down on a lot of repetition, because of certain patterns that will always occur. To see how this works it is best to look at an example.

Consider the following long division problem.

Here it is done in the usual way.
Notice there is a lot of bare space and a lot of repetition. For example most of the coefficients in the final answer occur 3 times. First we can get rid of the x in the divisor and all the results of multiplying by it, because we know that they will always subtract to 0.
Then as long as we keep things in the right order, we don't really need the variables. All of what we are doing can really be seen as operations on the coefficients and then at the end we can reconstruct the polynomial from its coefficients. If we do that, we get a division that looks like this.
In this form what we are doing is this. We get the numbers on the top by copying the first number from the bottom and then we multiply and then subtract and bring down. But there is still a lot of repetition and empty space. So lets see what we can do about that. If we get rid of the brought down numbers, it looks like this
and now we can collapse things to get this.
I did this with Photoshop by cutting and pasting, so now I need to clean up. Notice also that except for the very first number, the bottom row of numbers is the same as the answer row and the first number of the answer row is just the first coefficient of the divisor, so if we just copy that number down, we can get rid of the top row and just use the bottom row for our answer.
Now the procedure goes like this. Bring down the first coefficient and multiply it by the constant of the divisor, in this case the 3, and put the answer below the next coefficient. Then subtract and multiply and keep subtracting and multiplying until you get to the end. All the numbers in the bottom row except for the last one represent the coefficients of the answer, and the last one is the remainder, so by putting these numbers on a polynomial we get that our answer is x4-2x3+8x2-19x+54 R -156.

But there are a couple more simplifications we can make to make it even easier. As for notation I find it quicker to simplify the division sign because since there is nothing on top anymore there doesn't need to be a line there, so I like to write it like short division like this.

We can also do something to make the subtraction easier. It is easy to make mistakes with subtraction of plus and minus numbers, so you are probably going to do the subtraction by changing the signs of the second row and adding, like this.
There is a little trick we can do, though, to take care of changing the signs easier. We can change the 3 into a -3 and then multiplying by the minus number will automatically change them, so the final process looks like this.
If you just remember to always change the sign of the constant in the divisor then you can add instead of subtracting. So now the final procedure becomes this. Bring down the first number and then multiplying by the 'divisor' and put the answer under the second coefficient and then add and then multiply and then add until you get through the whole problem. Then just like before, use the first numbers in the bottom row and the coefficients of the answer, and the last number as the remainder. So in this problem you would do these steps. Then use the first numbers as the coefficients of the divisor and the last number as the remainder to get
x4-2x3+8x2-19x+54 R -156
for the final answer. Just one thing that you have to be careful about with synthetic division. If you have a polynomial with missing terms that you are dividing into you have to make sure to put zeros for the missing terms. So for example if you want to divide x4-2x2+3x+4 by x-2, you would write it as
.
Notice also that because now this is x-2 we use 2 instead of -2 the same way we used -3 instead of 3, so that we can add instead of subtracting. 

Using Synthetic Division

Now if you can guess a factor of this form for a polynomial you can easily check it by doing synthetic division. If the remainder is zero then you know it is a factor and then the numbers at the bottom will represent the coefficients of the other polynomial in the factorization. Then once you have the polynomial factored this way you can use the principle of zero products and set each factor equal to zero if you want to find the roots to the polynomial. It may not be so obvious how to solve the equation you get by setting the other factor equal to zero, but at least will have reduced the degree by 1, so it should be easier to solve than the original one. Sometimes what you can do is find enough solutions to reduce the degree down to 2 and then solve by the quadratic formula.

Notice that when 0 is the remainder when dividing by x-r then x-r is a factor, so x=r is a solution, so at least in this case the remainder when dividing by x-r is the value of the polynomial at r. It turns out that this generalizes and it is always the case that the remainder when dividing by x-r will be P(r). This result is called the remainder theorem, and here is why it works. By the checking method for division,

P(x)=(x-r)Q(x)+R,

so in particular

P(r)=(r-r)Q(r)+R
=0Q(r)+R=R.

Because of this synthetic division can also be used as synthetic substitution, an easier method for substituting a number into polynomial. To find P(r) by synthetic substitution you just do synthetic division with x-r and the remainder will be your answer.

There is a curious and rather nice thing about this, though. The signs change twice, which means not at all. Remember for synthetic division we change the sign of the number in our divisor so that we can add instead of subtract? Well, when substituting r, we change the signs again, because we are dividing by x-r. So to substitute r, we do synthetic division with -r, but then for the synthetic division, to change subtraction into addition, we change the -r back to r, so the end result of this is that for synthetic substitution you don't have to change the signs at all. You just use the number you are given.

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