## Saturday, June 28, 2014

### Algebra: Dividing Polynomials, Part 1 of 2

Dividing polynomials may appear more difficult to the student who is either unfamiliar/unpracticed in regular long division or is not comfortable with the rules of exponents. This topic will be covered in a two-part series. Today, we'll cover the basics of long division with numbers and a very simple polynomial division problem. In part 2, we will review three more complicated types of problems you may encounter.

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 subtract. We get 1 and then we "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 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!