Algebra: Dividing Polynomials, Part 2 of 2

In Part 1 of this series, I explained standard long division of numbers, and we covered a very basic polynomial division problem. Today, we will cover three situations you may encounter: (1) when there is a "missing term" in the dividend, (2) when the leading co-efficient of the dividend is not 1, (3) when there is a remainder.
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Situation #1: How to set up the problem when there is a "missing term" in the dividend.

Look at the problem above. Do you see the where the "missing term" should be? If you look at the exponents of the dividend, you'll see that there is no term for x^2. In order to properly divide these polynomials, we need to reintroduce the missing term. We can do that without changing the value of the dividend by using a co-efficient of zero, as shown circled in red below.

Now that the problem is set up in a familiar way, you can solve it as we did the problem in the Part 1. The steps are shown in the image above.

You should be aware that there may be multiple "missing terms", as in the polynomial (x^5 - x^2 + x). In such a case, introduce enough terms that all exponents are covered. In this example, you would write (x^5 + 0x^4 + 0x^3 - x^2 + x + 0) under the division bar.
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Situation #2: The dividend's leading co-efficient is not 1.

The leading co-efficient in a polynomial is the co-efficient on the term with the highest exponent. In the polynomial (2x^3 - x^2 - 3x + 2), the leading co-efficient is 2 because it is on the term with the highest exponent, 3. All of the problems we have dealt with so far have had a leading co-efficient of one. Let's use the polynomial I've given you already to work another division problem.

The first step of this problem is just like the others we have solved. Begin by dividing the first term of the dividend (green circle) by the first term of the divisor (orange box) to get the first term of the quotient (red box). Then, multiply this first term of the quotient by the whole divisor (underlined in purple). This new polynomial (underlined in blue) is written below the dividend and subtracted. Do you see how this step is just like the first step of the other problems we solved? The only, very minor difference, is the initial division of the first terms of the dividend and divisor.

Now, you can finish solving the problem as you would any other, shown below.

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Situation #3: The remainder is not zero.

Consider the problem I have worked below. You will notice that at the end, I am left with (-5x+1) (red box). I cannot divide (-5x) by the first term of the divisor, (x^2) because (x^2) is larger. What do we do with this remainder?

Your instructor should explain how to represent this remainder. Two representations are given below, although the second looks more professional and is my preference.

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I hope this series has been helpful to you. If you have any questions about this topic or something you would like me to cover in the future, leave them below. Thanks!