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Math 101: College Algebra12 chapters | 95 lessons | 11 flashcard sets

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Lesson Transcript

Instructor:
*Kathryn Maloney*

Kathryn teaches college math. She holds a master's degree in Learning and Technology.

Let's look at some more polynomial division problems. We will use long division and synthetic division, but this time we will have a couple of more involved problems. So, get out some paper and a pencil and let's begin!

We have looked at some basic polynomial long division and synthetic division. Now, let's have a look at a couple more complicated problems. They work exactly the same, except we have larger divisors or parts of the dividend are missing.

Here's our first: (*x*^3 - 9) ÷ (*x*^2 + 3).

Did you notice that we don't have an *x*^2 or *x* term in the dividend? We need to put something in as a place marker. Whenever we have a missing term, we will put in 0 as their coefficients. For example, in this problem we will have (*x*^3 + 0*x*^2 + 0*x* - 9). Why zeros? Because anything multiplied by zero is zero, so we aren't adding anything to the problem!

To fill in our long division, we now have (*x*^3 + 0*x*^2 + 0*x* - 9) as the dividend, so it goes under the long division symbol. The divisor is (*x*^2 + 3). Do you notice it's also missing a term? Just like the dividend, we need to put in a place marker. So, (*x*^2 + 0*x* + 3) goes to the outside, in the front of the long division symbol.

Now, we're ready to start the division. The steps are the same every time. It's easier to show and talk about the steps than just listing them. To find the first term of the quotient, we take the first terms from the divisor and dividend and then divide them: *x^3* ÷ *x*^2. I like to write them as a fraction. It's easier to divide or reduce that way: *x*^3/*x*^2 = *x*.

*x* is the first term of the quotient, so we'll write it above the long division symbol. To figure out what we will subtract from the dividend, we multiply *x* times the divisor, (*x*^2 + 0*x* + 3): *x*(*x*^2 + 0*x* + 3) = *x*^3 + 0*x*^2 + 3*x*.

*x*^3 + 0*x*^2 + 3*x* is written under the dividend, matching like terms. Here's where I like to make life easy. To subtract the polynomials, I just change the signs of its terms and add:

*x*^3 + -*x*^3 = 0- 0
*x*^2 + -0*x*^2 = 0 - 0
*x*+ -3*x*= -3*x*

Just like we did in long division, we bring down the next term, which is -9. Now, we have a predicament: *x*^2 doesn't go into -3*x* - 9! This turns out to be our remainder. What's our answer then? *x* + ((-3*x* - 9)/(*x*^2 + 3)). There isn't a need to put 0*x* as part of the denominator in the remainder.

Here's another example: (2*x*^3 + 10 - 14*x*) ÷ (*x* + 3).

This one is almost ready for synthetic division. The divisor is a first-degree binomial with a leading coefficient of 1. Place the coefficients of the dividend under the symbol, just like in long division, but do not write the variables! Make sure to leave space in between each number. You don't want to get them confused.

Wait! Did you write it as you see it? Oh no! The dividend and divisor must be in descending order. What does that mean? We need to list the terms from the largest exponent to the smallest. Even if we're missing one, we need to put it in as a place marker. So, now we have (2*x*^3 + 0*x*^2 - 14*x* + 10) ÷ (*x* + 3).

Next, look at (*x* + 3). We write -3 to the left of the long division symbol. Remember, you always take the opposite of what you see in the divisor. Now, we're ready to start synthetic division.

Our first step is to bring down the leading coefficient, 2. Multiply -3 times 2 and place the result underneath the next coefficient, 0. So, we have -6 written underneath the 0. Add 0 + -6, and write that next to the leading coefficient, 2.

Multiply -3 times -6 and place the result under the next coefficient, -14. So, we have 18 written under the -14. Add -14 + 18, and write the sum of 4 next to the -6.

Multiply -3 times 4, and place the result underneath the next coefficient, 10. So, we have -12 written underneath the 10. Add 10 + -12, and write the sum of -2 next to the 4. Here is our final answer with the remainder:

How do I know there's a remainder? Because our last term is not a 0.

But, what is the answer going to look like? In synthetic division, the degree of the final polynomial answer is one less than the dividend polynomial. Since 2*x*^3 + 10 - 14*x* is degree 3, our answer will be degree 2.

Starting from the left, we will have 2*x*^2 - 6*x* + 4 with a remainder of -2. The remainder will be written the same as if we had done this problem as long division - a fraction. So, our answer will look like 2*x*^2 - 6*x* + 4 + (-2/(*x* + 3)).

Here's our next example: (6*y*^3 + 7*y*^2 - 5*y* + 5) ÷ (3*y* - 1).

To fill in our long division, (6*y*^3 + 7*y*^2 - 5*y* + 5) is the dividend, so it goes under the long division symbol. (3*y* - 1) is the divisor, so it goes to the outside, in the front of the long division symbol. Now, we're ready to start the division.

To find the first term of the quotient, we take the first terms from the divisor and dividend and divide them: 6*y*^3 / 3*y*. I like to write them as a fraction. It's easier to divide: 6*y*^3 / 3*y* = 2*y*^2. 2*y*^2 is the first term of the quotient. We write it above the long division symbol.

To figure out what we will subtract from the dividend, we multiply 2*y*^2 times the divisor, 3*y* - 1. 6*y*^3 - 2*y*^2 is written under the dividend, matching like terms. To subtract the polynomials, I just change the sign of the terms and add.

- 6
*y*^3 - 6*y*^3 = 0 - 7
*y*^2 + 2*y*^2 = 9*y*^2

Just like we did in long division, we bring down the next term, which is -5*y*. Guess what? We do the same exact step. To find the next term of the quotient, we take the new first terms and divide them: 9*y*^2 / 3*y*. I like to write them as a fraction: 9*y*^2 / 3*y* = 3*y*. 3*y* is the next term of the quotient. Remember, that gets written next to the 2*y*^2 above the long division symbol.

To figure out what we will subtract from the dividend, we multiply 3*y* times the divisor, 3*y* - 1. 3*y*(3*y* - 1) = 9*y*^2 - 3*y*. 9*y*^2 - 3*y* is written under the dividend, matching like terms. To subtract the polynomials, I just change the signs of the terms and add:

- 9
*y*^2 - 9*y*^2 = 0 - -5
*y*+ 3*y*= -2*y*

Just like we did in long division, we bring down the next term, which is 5. You'll notice that the dividend is now -2*y* + 5. When we divide to find the next term of the quotient, we take the new first terms and divide them. -2*y*/3*y*. Do you see we end up with a fraction, -2/3? Well, we don't like fractions in the quotient except as a remainder, so -2*y* + 5 turns out to be our remainder. This is what the final answer will look like:

2*y*^2 + 3*y* + ((-2*y* + 5)/(3*y* - 1))

Before we end this video, let's review when to use long division and when to use synthetic division. My students really like synthetic division, but it isn't right for every division problem. How do I know the difference?

Any division problem that has a divisor as a one-degree binomial is perfect for synthetic division. That's why we used it on (2*x*^3 + 10 - 14*x*) ÷ (*x* + 3). *x* + 3 is a one-degree binomial with a leading coefficient of 1.

Okay. It is true we can use synthetic division with a divisor that does not have a leading coefficient of 1, but that'll give us a fraction. Fractions are messy, and students tend to make more mistakes when using them, so keep to a one-degree binomial with a leading coefficient of 1.

Any other division problem needs to be done using long division. That's why we used it on (6*y*^3 + 7*y*^2 - 5*y* + 5) ÷ (3*y* - 1) and (*x*^3 - 9) ÷ (*x*^2 + 3).

One final rule to keep in mind as you do polynomial division: always make sure your polynomials are in descending order, largest exponent to smallest. If you're missing a term, put a zero to hold that place, like we did in example #1.

Following this lesson, you'll be able to:

- Determine when to use long division versus synthetic division for polynomials
- Explain the rules for both types of division
- Work problems using both types of division

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Math 101: College Algebra12 chapters | 95 lessons | 11 flashcard sets

- What Are the Five Main Exponent Properties? 5:26
- How to Define a Zero and Negative Exponent 3:13
- How to Simplify Expressions with Exponents 4:52
- Rational Exponents 3:22
- Simplifying Expressions with Rational Exponents 7:41
- How to Graph Cubics, Quartics, Quintics and Beyond 11:14
- How to Add, Subtract and Multiply Polynomials 6:53
- How to Divide Polynomials with Long Division 8:05
- How to Use Synthetic Division to Divide Polynomials 6:51
- Dividing Polynomials with Long and Synthetic Division: Practice Problems 10:11
- Go to Exponents and Polynomials

- Go to Functions

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