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Math 101: College Algebra12 chapters | 94 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.

Adding, subtracting and multiplying polynomials are, basically, the same as adding, subtracting and multiplying numbers. They only difference is that we have a pesky variable to worry about, but this video will show you that's no problem, so no worries! This method has worked for many of my students, and I think it will work for you, too!

Adding polynomials is very easy! There are many ways to add polynomials that have been taught, but here's my favorite. I look at the problem and put a circle, square or triangle around the like terms so I don't get them confused.

Let's look at this example.

(2*x*^2 + 3*x* + 4) + (*x*^2 - 5*x* + 7)

I like to start from the left. I circle 2*x*^2. Now, I look for another like term to 2*x*^2. Of course, that's *x*^2, so I circle it.

Then I add them: 2*x*^2 + *x*^2 is 3*x*^2, and then I write that down as part of my answer.

The next term is 3*x*. I put a square around that one. Then look for another one. Of course, we have -5*x*, so I put a square around that one too.

Then add them: 3*x* + -5*x* is -2*x*, and I write that down as the next part of my answer.

You always want to double check that there aren't any more.

Finally, we have 4. I continue the same idea, but I put a triangle around it. Now I look for another like term to 4. Of course, that's 7, so I put a triangle around that one.

Then I add them: 4 + 7 is 11, and I write that down as the last part of my answer.

It's always good to double check that you have everything in a circle, square or triangle. When you get longer polynomials, it's easy to miss terms!

So here's my final answer: 3*x*^2 - 2*x* + 11.

'Okay, Kathryn, do I have to use circles, squares and triangles?' No! If you're good at adding polynomials, you can cross off as you go, but for those that haven't had enough practice, this is definitely my suggestion.

'Kathryn, what about problems that have more terms? What other shapes or ideas do you have?' Great question! You can use different colors to circle like terms. I don't use it here for the benefit of students who are color blind, but you could have used red to circle the *x*^2s, blue to circle the *x*s and green for the numbers. I've also used the method of underlining like terms too! Whatever you choose is how you will distinguish the different terms!

In subtraction, let me show you the underlining method.

(3*x*^2 - 2*x* + 5) - (2*x*^2 - 6*x* + 7)

First, I am going to distribute the -1 into the second expression. That will make this an addition problem!

The first expression stays the same: 3*x*^2 - 2*x* + 5. We will distribute the negative, like this: -1 * 2*x*^2, which is -2*x*^2; -1 * -6*x*, which is a positive 6*x*; and -1 * 7, which is -7. This gives us our new subtraction problem: (3*x*^2 - 2*x* + 5) + (-2*x*^2 + 6*x* - 7).

Remember, I am going to show you the method of underlining instead of circling to add the expressions.

We look at the first term 3*x*^2 and underline it. Now, I continue to look for a like term. Here it is, -2*x*^2, and I underline it. Now I add them: 3*x*^2 + (-2*x*^2), and we get *x*^2. That's going to be the first term of our answer.

The second term is -2*x*, and this time, I put a squiggly line under it. Now I continue to look for a like term. Here it is: 6*x*, and I put a squiggly line under that one. I add -2*x* + 6*x*, and I get 4*x*. That's the second term in our answer.

The last term is 5, and this time I put a jagged line under it. Now, I continue to look for a like term. Here it is: -7, and I underline it. Now, I add them: 5 + -7, and we get -2. This will be my last term in the answer: *x*^2 + 4*x* -2.

Now, we get to multiplication. This problem won't quite work like addition or subtraction, and we can't use FOIL because these are larger than a binomial times a binomial!

(*x*+5)( *x*^2+3*x*-2)

First, multiply the first term in the product: *x* times everything in the second expression.

I like to draw arrows to remind me which multiplication I've done; otherwise I tend to get lost.

This is how it will look:

*x*(*x*^2) + *x*(3*x*) + *x*(-2)

Let's multiply.

*x*(x^2) = *x*^3 + *x*(3*x*) = 3*x*^2 + *x*(-2)= -2*x*.

This isn't our final answer; we need to multiply everything in the second expression by 5!

So we'll have 5(*x*^2) + 5(3*x*) + 5(-2)

Are you ready for the final answer? We simply add the like terms together!

*x*^3 + 3*x*^2 - 2*x* + 5*x*^2 + 15*x* - 10

Start from the left, and circle *x*^3. It looks like there aren't any like terms for *x*^3, so we write that down as our final answer.

Put a square around 3*x*^2. I look and find 5*x*^2, so I put a square around that term, too. I don't see any more, so 3*x*^2 + 5*x*^2 = 8*x*^2. 8*x*^2 is written next to *x*^3 as part of our final answer.

Put a triangle around -2*x*. I look and find 15*x*, so I put a triangle around that term too. Why? Well, they're like terms. I don't see any more like terms for -*x*, so -2*x* + 15*x* = 13*x*. 13*x* is part of our final answer, and I'm going to write it next to 8*x*^2.

Finally, I see -10. I underline this term and look for another one like it. I don't see one, so -10 is written in my final answer.

So, what is the final answer then?

*x*^3 + 8*x*^2 + 13*x* - 10.

Adding: No trick here. Just add like terms together. If you get lost at first, use circles, squares, triangles, lines...anything to separate the different terms.

Subtraction: Multiply the second expression by -1, and add the two expressions together.

Multiplication: Multiply each term in the first expression by each term in the second expression. Add like terms together and you're done!

By the end of this lesson you'll know all about adding, subtracting and multiplying polynomials.

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Math 101: College Algebra12 chapters | 94 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 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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