If you’re having trouble with this subject, you’ve come to the right place. If you don’t know what this is, and you’re curious, this is also the right place for you! Polynomial division is fairly important, and is used throughout calculus.
What is polynomial division?
Polynomial division is when you divide one polynomial by another. A polynomial just means an expression where there are multiple powers of x. Here are a couple of examples that require polynomial division.
Long Division
Long division is really important to know, because it can be used to divide any two polynomials. There are no exceptions or rules to this. Synthetic division is a nice trick to know, but it is only for a single type of polynomial division. Long division covers them all. Let me set up a problem to show you the technique of long division. I’ll explain it all, step by step. For some reason, you would like to do the division in this expression:
First, we need to write this expression in long division form. You write it the same way you do normal long division, as if they are only simple numbers. I’ll show you what I mean very soon. There is one catch to this. You need to use place holders if you have powers of x that are missing. The dividend, the thing on the inside being divided, needs to have all powers of x represented in decreasing order. The first term must be of the highest power of x that is in the expression. The second term must be one lower power of x. If there is none of that power, you put a zero. In this first example, there is no x². I’ll show you how it’s done.
This is the correct format that needs to be used when doing long division. So, what is the first step? We need to look at the divisor, the thing outside of the box, on the right side. In this case, x – 2. In the divisor, you look only at the highest power of x. In this example, we have only an x. The coefficient is just 1. So, we look at the x. What times ‘x’ is equal to the first term of the dividend? In other words, what times x equals x³? This is pretty simple, and the answer is obviously x². This is the first term of the quotient (our answer). Write this on top of the line, exactly above the first term of the dividend.
Now, like in normal long division, we multiply this first part of our answer by the divisor, the expression on the right side. Then, you subtract that from the dividend! After this subtraction, you bring down the next term in the dividend, and repeat the process. I know this seems complicated when you are reading it for the first time, but I’ll show you what I mean.
See how I did that? Now we repeat the process until the first term of the divisor can no longer “go into” the term we are looking at in the dividend. After that, I’ll explain what we do. For now, look at the first term of the divisor, the x. Of this new expression, 2x² – 5x, look at the first term. What times x equals 2x²? Then we just continue doing what we just did. I’ll complete the rest of the process.
Now, I’ve run out of terms in the dividend. There is nothing more to “bring down”, therefore there is nothing more that x can “go into”. I must stop dividing, and reveal my answer. The solution to the division is everything that is on the top line. That is your answer. However, there is one more part that we cannot forget. This is exactly the same as regular long division. When you run out of things to bring down, you must look and see what your remainder is. You remainder is whatever is left over after you do the last subtraction in the long division. Here, we have 0 left over. This means we do not have a remainder. If you have a remainder, you must adjust your answer. You need to add one more term to the expression on the top line. This term is the remainder divided by the divisor. If we had a remainder of 3, for example, we would add 
.
The final answer can be written like this.
I’ll do one more example for you.
There you go! This technique can be used to divide any two polynomials. Try it out, as it can be useful in several situations in calculus!
Synthetic Division
Alright. You are proficient in long division now, right? You’ve got that down? Now you want a trick? Is that what I’m supposed to do now? Okay, well I have one for you. This is called synthetic division. Now, it only works in a specific situation, but it can be extremely useful sometimes. If you don’t believe me, check out this post where I needed it to solve the problem. I actually give the full explanation of synthetic division in that post, but I’ll do it again here just so you guys can see it together with long division. The post was a question submitted by a reader, asking about how to take the limit of something fairly difficult. Click here to go right to the post and see how I used synthetic division.
So, when can we use synthetic division? A condition must be met in order to use it. We need to look at the divisor. This is the expression we are dividing by. This needs to be a single x plus or minus a constant number. This could be x – 3, or it could be x + 10. It doesn’t matter if it is a plus or if it is a minus. It just needs to be one of these simple forms. Let me show you how it is done through an example.
Now, there are a few similarities to long division. First, you need to use place holders in the exact same way. If you don’t have any x² term like in the example above, you need to put a 0. Also, the last number on the bottom is the remainder, and is treated the same way as it is in long division. The remainder over the divisor is added to the solution, in order for it to be correct. If the remainder is zero, you can ignore this step. Okay, so below I will show you the way to set up synthetic division. First, you must think of the divisor, the x – 3, always in this for. x – a. a can be anything, even a negative number. If the divisor were x + 5, a would be -5, right? Place -3, the a, in the upper right hand corner. Then, in a straight row, right the coefficients of each term. If there is zero of a term, make sure to write 0. Below is the proper way to set up the division. Leave room for a second row under the first, and draw a line under that space.
To begin, bring the first term in the top row straight down.
Next, multiply the number in the top right corner by this number you just brought straight down. In this example, multiply the 3 by the 1. Write this number in the second row underneath the next number of the first row.
Add the two numbers that are now right above/below each other, and write this number underneath them.
Now, repeat the process again. Multiply the upper right hand number by this new bottom number. Add that to the next number of the top row. And repeat until there are no more numbers in the top row. The final number in the bottom row is the remainder. Place this in a box, and leave it off to the side where it belongs.
Not including the remainder, these are your new coefficients! The new polynomial has the highest power of x that is one less than the highest power that was in the dividend. In this problem the original highest power of x was x³, so the new highest power will be x². Just place on the coefficients, and add the remainder if necessary. This is the solution!
I believe that I explained this better in the solution I posted to a reader’s question. I encourage you to check out that page if you have any remaining questions. Click here if you’d like to view that post. I’ll show you one more example.
