First, let's review regular long division. If you are already familiar with this, you can skip ahead to the next portion of this post. Unfortunately, many kids get through third grade (or whenever they teach this now) without a good handle on the division process. Then, when those kids get to Algebra class, they struggle Big Time with dividing polynomials. The great news is that it is never too late to learn AND this is a useful skill, even in Calculus!

Let's define some terms. In a division equation, the number being divided is called the

**dividend**. The number that is dividing is the**divisor**. The result of dividing the*dividend*__by the__*divisor*is called the**quotient**. If the*divisor*does not divide evenly into the*dividend*, there will be a**remainder**, which is usually expressed as a fraction.
Below is a typical long division problem being worked in steps. The explanation is given below the image.

We divide the

*divisor*into the first number of the*dividend*(red circles). In this case, we say "2 goes into 5, 2 times." So, we write the step 1*quotient*(bold circle) above the number in the*dividend*we used: we write 2 above the 5. [NOTE: When dividing polynomials, the*divisor***will never be larger than**the first "term"/"place" of the*dividend*.] Then, we write the**product**of 2 and 2 (4, red square)__below__the 5 and**. We get 1 and then we**__subtract____"bring down"__the next place in the*dividend*, in this case, a 0, making 10.
In step 2, our new

*dividend*is 10 (green circle). We divide and see that "2 goes into 10, 5 times." So, we write the step 2*quotient*(bold circle) above the next place of the original*dividend*(above the 0, which is what we "brought down" to make 10). Then, we write the product of 2 and 5 (10, green square), below the 10 and subtract. We get 0 and then bring down the next place in the original dividend, which is a 4.
We continue this process, which takes about 5 steps, seen in the image above, until we have a number left that is

*not divisible*by our*divisor*. That is, 1 is less than 2. Therefore, we call 1 our**remainder**(bold blue circle). We write the*remainder*(blue circle) next to the*quotient*and show it over the*divisor*(purple circle) as a fraction. The entire number 25,215 1/2 (bold pink box) is the answer (*quotient*) to the division problem.
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We can apply these same general steps when we divide polynomials. You should be familiar with exponent rules. If you aren't, check out this website that will give you a good overview.

Let's begin with a pretty basic example. Divide (x^3 + 5x^2 + 7x + 2) by (x + 2). After we set up the problem, our first step is to divide the

Then, we multiply the first step

You can see step 2 in the image below. This time, we divide the first term of the

Look at the image of step 3 below. It follows steps 1 and 2 exactly. Can you follow along?

After subtracting at the end of step 3, we are left with a

You've just walked through a basic polynomial division problem with me. Join me next time for part 2 of this series where we will look at the following situations: (1) when a term is "missing," (2) when the leading co-efficient is not 1, and (3) when there is a remainder.

As always, feel free to leave any questions in the comment section below. If you have ideas for future posts, feel free to leave them here as well!

We can apply these same general steps when we divide polynomials. You should be familiar with exponent rules. If you aren't, check out this website that will give you a good overview.

Let's begin with a pretty basic example. Divide (x^3 + 5x^2 + 7x + 2) by (x + 2). After we set up the problem, our first step is to divide the

**first term**of the*divisor*(blue box) into the**first term**of the*dividend*(blue circle). Here, that means we are dividing x^3 by x. According to the rules of exponents, if the base is the same (x is the base here), you simply*subtract*the exponents. Since 3 - 1 = 2, x^3 divided by x is x^2. We write this first step*quotient*(red circle) above the first term of the*dividend*(blue circle).Then, we multiply the first step

*quotient*(red circle) by the entire*divisor*(underlined in green). We get x^3 + 2x (underlined in orange), which we write__below__the*dividend*. Then, we__subtract__terms. I suggest that you write (x^3 + 2x) in parentheses as I have done so that when you subtract, you remember to subtract both terms. This will be helpful when some of the terms may be negative (we will cover that in the next post of this series). After subtracting, we have 3x^2. We then bring down (arrow) the next term of the*dividend*(**with its sign)**to set up step 2.You can see step 2 in the image below. This time, we divide the first term of the

*divisor*(blue box) into the first term of the new*dividend*at the bottom (blue circle). That is, divide 3x^2 by x, which yields 3x (red circle). Then multiply this second step*quotient*(red circle) by the entire*divisor*(underlined in green), yielding the product (3x^2 + 6x), underlined in orange. We write this below the second step*dividend*we used, and__subtract__terms. This leaves us with (x + 2) and we can move onto step 3.Look at the image of step 3 below. It follows steps 1 and 2 exactly. Can you follow along?

After subtracting at the end of step 3, we are left with a

*remainder*of zero. That means our*divisor*divides**evenly**into our*dividend*, and our final*quotient*is x^2 + 3x + 1.You've just walked through a basic polynomial division problem with me. Join me next time for part 2 of this series where we will look at the following situations: (1) when a term is "missing," (2) when the leading co-efficient is not 1, and (3) when there is a remainder.

As always, feel free to leave any questions in the comment section below. If you have ideas for future posts, feel free to leave them here as well!