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## Multiplying Polynomials

In the last lesson, we learned how to add and subtract polynomials. Let's continue building on that knowledge and try to multiply polynomials. You'll still need to remember how to work with exponents and the idea of like terms. If any of those gave you some trouble, I would go back and do some more review. Before we multiply polynomials, let's start with something a little bit more simple. Let's try an area problem.

## Area and Multiplication

We're going to use the idea of area to help us understand multiplication and multiplying polynomials. If I want to represent the multiplication of 10 times Multiplying 10 and 12 would give me the total area of the square, that would be incredible next. We're going to split up 12 into 10 plus 2. I can also set up another rectangle. The width will be 10, and the length will be 10 plus 2. Notice I've created 2 smaller rectangles that make up my larger rectangle. So what's the area of each of these rectangles? Write your answer here.

## Area and Multiplication

This rectangle would be 100 square units, and this one would be 20 square units, we know this because 10 times 10 is 100, and 2 times 10 is 20. If this side of the rectangle is 10, then this side of is 10 and this side is 10. And here's the amazing part, even though I split up one of the sides of the rectangle, my total area is still the same, 120 and 120. We simply add up these two smaller areas to get the total area of a rectangle.

## Area and Factors

We work with numbers, so of course, what are we going to do next? Let's work with variables. The variable just holds the place of a number. So this x could just be like my 10. Here's another rectangle to represent this expression. One side or one factor is x, and the other factor is x plus 2. The sides of this rectangle represent the factors in this expression. Factors are simply numbers that are multiplied together. Here's the first number x and here's the second number x plus 2. We know each of these is just a number because x represents a number. So, here's one number this would be a different number. Using what we saw before, what do you think each of these areas are?

## Area and Factors

This area would be x squared. And this one would be 2x. Great job if you got both of those correct. This shape has one side of x and another side of x. So it really makes it a square. So the area has to be x times x, or x squared. For this rectangle, we know one side is 2 and this other side is x, because this side is x. So we have 2 times x, or 2x.

## More Area and Factors

You might be wondering what this rectangle has to do with this expression. Well, really the multiplication of this factor, and this factor, represents the total area inside a rectangle. We know this answer should be x squared and 2x, or x squared plus 2x, you would have to add these two areas together to get the total area. If we distribute the x to each term on the inside of the parenthesis. We would have x times x, which is x squared, and x times 2 which is positive 2x. If we list the sum of the terms to represent the total area, we've really found the multiplication of 2 factors. Some students have trouble remembering x times x is x squared, sometimes I think it's 2x. But we know that can't be the case because if we have x by x or x times x, we formed a square or x squared. I hope this is a powerful visual for you to remember x times x is always x squared.

## Commutative Property in Area

The last thing I want to mention about this problem is what if, instead, we had x plus 2 times x? If we take our rectangle and rotate it 90 degrees we can see that our factors are just along different sides. We have one factor with a side of x plus 2 and the other factor with a side of x. Notice that we would get the same total area, x squared plus 2x. Again, this is another demonstration of the commutative property of multiplication. The order of the factors doesn't matter.

## Multiply by a Monomial

Let's try multiplying this binomial by a monomial. Use what you know about multiplication and exponents to figure it out. We haven't done one like this before, so don't worry if you don't get it right on your first try. Always try to reason through it, and work from what you know.

## Monomial Multiplication

Use what you know about standard form and multiplying polynomials to answer this question. What's the product of this monomial with this binomial? Write your answer here.

## Monomial Multiplication

Negative 10y to the 4th plus 8y cubed is the correct answer. That's some great work if you got it correct. If you got stuck somewhere, stay with me and then pause it when you find a mistake and then keep going. This question involved exponents and a negative term, so there are a couple of places where you might have gotten tripped up. I'm going to list this factor, 2y squared, on one side of a rectangle, then I'm going to have 4y, and negative 5y squared on the other side of my rectangle. I'm going to consider this a positive region, because I have a positive number times a positive number. I shaded this region orange, because we're going to wind up with a positive number, here's a positive times a positive. For this region I've shade it grey because we have a negative number times a positive number, we'll wind up with a negative area here. 2y squared times positive 4y is 8y cubed. Next I have 2y squared times negative 5y squared, I'm finding the area of this part of my rectangle. I know a negative area might not make sense, but you can think about this area as the area that you have and this area, about the area that you owe. We use our properties of numbers to multiply 2 and negative 5, and our properties of exponents to get negative 10y to the fourth. Finally, we reverse the order or the two terms, to get our final answer. Negative 10y to the 4th should be our first term, and positive 8y to the them.

## Multiply Two Digit Numbers

Now that we've covered monomials, let's make this a bit more challenging. Let's multiply two binomials. I'm going to start with numbers to give a sense of the model. Now, even I can't do this that quickly in my head. I actually go about this another way. I split up 12 into 10 plus 2. And 14 into 10 plus 4. When I do first factor, or side of my rectangle is 10 and 2. The second factor, or side of my rectangle, is 10 plus 4. Now this is pretty easy to visualize and think about multiplying the numbers together. I would have 100, 40, 20 and 8. To get each of these areas, I'm simply multiplying the sides of the rectangles. This one is 10, and this one is 10, so 100. This rectangle is 2, and this side is 10 because this side is 10, so 20. Here, this opposite side is also 10, so I have 10 times of 8. Adding up all of these areas, I get 168. Try multiplying this out and you'll find it's 168. Pretty amazing. This also makes sense because I have 10 times 10, which is 100. These 1s are really in the tens spot. Then I have 2 times 10 which is 20. Again, this 1 is in the tens spot. 4 times 10, which is way to think about multiplication for some of you.

## Multiply Two Binomials

So, let's try something harder. What do you think the area of these rectangles would be? I've gone ahead and listed the factor of x plus 2 on one side of the rectangle and x plus 4 on the other.

## Multiply Two Binomials

If you got x squared, 4x, 2x and 8, excellent work. I know I have a square of x on each side, so the area must be x squared. For this rectangle, one side is x and the other side is 4. So I get an area of 4 times x. This rectangle has an area of 2x, since this side is 2 and this side is x. Our last rectangle has an area of 8. This short side is 2. And this horizontal piece is 4. 2 times 4 makes

## Total Area is a Product

Let's add up the area of all our rectangles together. What's the total area or the product of x plus 2 times x plus 4? Write your answer here.

## Total Area is a Product

The total area would be x squared plus 6x plus 8. I have like terms of 2x and can see the connection between 168 and this polynomial? How do these two relate to each other? As a hint, think about the place value of each of these numbers. And what do you think x could be. If you have some ideas, post in the forum.

## Multiply Binomials Negative Terms

Try this one out. What do you think would be the product of these two binomial? You may have learned how to multiply binomials in other ways, and you can use those if you'd like. If you're stuck or don't have a starting point, I suggest the rectangle.

## Multiply Binomials Negative Terms

Here, the product was 2x squared minus 1x minus 6. If you got this, you're quite the pro at multiplying binomials. Now, I know I threw a negative sign at you. So it's okay if you got stuck. There's also a coefficient of 2 here. So that might have thrown you off. I'm going to start by listing the factors on each side of my rectangle. I'll have 2x plus 3 and x minus 2. So I'll have 2x times x, which is 2x squared. Then I'll have 2x times negative 2, which is negative 4x. Notice that we're distributing or multiplying the 2x in the first parenthesis to each term in the second parenthesis. That also happens in our rectangles. Then I'll do 3 times x, which is 3x. And finally by making a little bit more room, I'll have 3 times negative 2, which is negative 6, our last area of our rectangle. And finally, we combine our like terms, the negative 4x and the 3x, to get our answer.

## Multiple Variables in Binomials

Now let's try something a little bit harder. Try multiplying these two binomials together. Notice that we have a variable in the first term and in the second term, in both binomials. Our process for multiplying works the same way, you want to be careful with your exponents and combine like terms. You can put your answer here, good luck.

## Multiple Variables in Binomials

We start with a rectangle. One side will be 2b minus 5c. And then this other side will be 9b plus 4c. Multiplying the first two terms together, we get 18b squared. Multiplying 2b times 4c, I get 8bc. Next, we distribute the negative 5c to 9b. Negative 5c times 9b is negative 45bc. And finally, we multiply negative squared minus 37 times bc minus 20c squared.

## Two and Three Digit Multiplication

We worked with multiplying monomials and binomials, and binomials with other binomials, now let's try something a little bit harder. We're going to multiply a binomial with a trinomial. Let's start with our simple case using numbers, and then see if you can extend that to figure out something harder. Just like before, we can split-up each of these numbers, so that way we can make this multiplication problem more simple, I'll have 13, which is 10 plus 3, and 123, which is 100 plus 20 plus 3. Now can imagine doing this multiplication using a rectangle with the appropriate side-lengths. When I find the area of each of these rectangles and add them up, I get 1,599, the answer to our multiplication problem.

## Binomial Multiplied by a Trinomial

In the case of multiplying a binomial by a trinomial, we just want to add some rectangles so we can complete our multiplication. Just like with these numbers, I'm going to set up a rectangle, and then I'm going to label each factor along the side of the rectangle. This one will be x plus 3, and this one will be x squared plus 2x plus 3. Using what you know about multiplication, I want you to figure out the area of each of these rectangles. What would they be?

## Binomial Multiplied by a Trinomial

For this first rectangle, we have x times x squared. So that's just x cubed. Remember, we add the exponents, 2 and positive 1. For this rectangle, we have x times 2x. So, the area's 2x squared. And this third rectangle over here is x times 3, or 3x. Notice, we're really just distributing this positive x to each of the 3 terms in the second trinomial. This leaves us with 3 new terms. Next, I have 3 times x squared, which is 3x squared. For this next rectangle, I have positive 3 times 2x, so I get 6x. For this last rectangle, we have positive 3 times positive 3 which is 9. The sides of the rectangle give us an area of 9.

## Total Area and Final Product

So find the product of this binomial multiplied by this trinomial by adding up the area of the rectangles. What's their sum?

## Total Area and Final Product

To find the product, we add up the areas of the rectangles. I've listed out all my areas, or my terms here, and I've listed the x squares together and the x's together. These are my like-terms, 3x squared plus 2x squared, makes 5x squared, and 6x plus 3x equals 9x, so here's my total sum. If you haven't made the connection before between this diagram and this diagram, we've simply replaced the 10 with an x. And in fact if we plugged in a 10 for x into this expression, we would get 1599, the same number here.

## Squaring a Binomial

Now that we know that we can use our area model in a bunch of different cases, let's see if we can use it here. We're going to try and square a binomial. When we square a binomial, we simply raise a binomial to the second power. And what we're wondering here is, is this equal to this? Let's use what we know to figure it out. I've drawn out the rectangle for you. Now I want you to fill in the products that you would get when you multiply each of the terms. Basically, you're finding the area for each of these rectangles. After you find the area of each of the rectangles, then write the sum, or the total area, in this box.

## Squaring a Binomial

If you got 4x squared plus 12x plus 9, excellent work. We know that first rectangle is actually a square, each side is 2x, so, 2x times 2x is 4x squared, this rectangle would be 2x times 3, which should be 6x. The same is true for this rectangle,, 2x times 3 makes 6x. And finally for this last rectangle, we would have 3 times 3, for each of these sides which makes 9. When we multiply 2x plus 3 by itself, we end up with this perfect square. This perfect square has an area of 4x squared plus 12x plus 9, which we know isn't equal to 4x squared plus from unit 1, we said that squares can't distribute over sums or differences. If we only squared the first term and the last term, we miss out on these middle terms of 6x, these are considered to be linear terms, they only have one variable with an exponent of 1. I now it's really easy just to think about squaring each of these terms to get this answer, but in this case, the math goes against our intuition, it's not true. This is what makes learning math interesting, and sometimes the patterns aren't exactly what they seem.

## Perfect Squares 1

Let's continue investigating these perfect squares. What I want you to do is multiply these 2 perfect squares out. What would you get here and here?

## Perfect Squares 1

Here are the correct answers. For a square, we know we need to multiply X plus 4 by itself twice, the same is true for 3 x plus 1. I draw out my rectangle and I label each of the sides. Finally, we add up all the rectanges to get our total area. I can only add the 4x and the 4x areas together because they're like terms. That's where I get 8x. Down in this square, we have 3x times 3x, which is is positive 1. We add up all of these areas just like before to get 9x squared plus 6x plus 1. Great work if you got these correct.

## Perfect Squares Pattern 1

Based on these two examples, maybe you can figure out a pattern. How did we get this answer from this term and this term? And how did we get this answer from this term and this term? Using what you see here, write a general rule for finding the quantity of a plus b squared.

## Perfect Squares Pattern 1

Well, to get this x squared, we multiplied x times x. And, to get this 9x squared, we multiplied 3x times 3x. So, we just squared the first term. If I squared the first term here I would get A squared. This middle term is a little bit trickier. If I do 4 times x I have 4x. But I don't need one of them, I need two. So I'm going to have 2 times my first term and my second term. The same is true down here. I don't just have one 3x, I have two. So I'm going to have the first term times the second term, times 2. So here's the first term, a, times the second term, b, times 2. I listed the 2 first, since we usually put coefficients in front of variables. Now we need to think about the last term, the 16 and the 1. Well I know 4 squared is 16, and 1 squared is 1. So to get the last term, all I need to do is square the second term. So we square the b, we add b squared. Here's a general pattern we could always use if we ever have to square a binomial. Now, you don't have to memorize this, but this is helpful in these particular cases. If you do commit this to memory, or continue to recognize the pattern, it will help you out in later lessons.

## Perfect Squares 2

Let's see if we can identify one more pattern using our perfect squares. This time we're going to have a subtraction sign in between the terms of our binomials. What I want you to do is start by squaring each of these binomials. What would the products be?

## Perfect Squares 2

If you got these 2 correct, excellent work. We'll set up our rectangles as usual. What we want to be careful about is working with these negative signs. I know x times negative 4 is negative 4x, so these 2 areas will wind up being negative. We also want to be careful when computing the last area. We have negative 4 times negative 4, which is positive 16. If we add up our like terms we'll have x squared minus 8 x plus 16. That's the total area or the product for the first binomial. If we square the second binomial we'll get 9 x squared minus

## Perfect Squares Pattern 2

Let's use these two examples to come up with a general rule for a minus b squared. What do you think this would be equal to?

## Perfect Squares Pattern 2

Well, to get the first term in both of our answers, we just squared the first term, x times x is x squared, and 3x times 3x is 9x squared. So again we're going to start off with a, a squared. We square the first term. This middle term is just like from before except instead of being positive they're negative. So I take the first term, multiply it by the second term, and then multiply it by 2. X times negative 4 is negative 4x times 2 is negative 8x. 3x times negative 1 is negative 3x times 2 is negative 6x. So, we take the first term,a, multiply it by the second term, negative b, and then multiply it by 2, and we get negative 2ab. This last part is simply this last term squared. Negative 4 times negative 4 is positive 16. Negative 1 times negative 1 is positive 1. So we know negative b times negative b is positive b squared. There's our pattern.

## Perfect Square Trinomials

The two patterns that we just found are actually called perfect square trinomials. They're called this because we can represent the binomial squared as a perfect square. When we multiply out the binomial, we'll get the trinomial, a squared plus 2ab plus b squared. The only difference between the two patterns is the plus sign and the minus sign, which also appears here in the second term. The second term is positive 2ab with the first one, and negative 2ab for the second. Let's see if you can use these patterns to answer our next question. What would you get be squaring this binomial? Write you answer here.

## Perfect Square Trinomials

Great work if you got 16x squared plus 40xy plus 25y squared. I start by listing out my factors twice. Then, I can draw a triangle to represent the areas that I'm going to calculate. 4x times 4x is 16x squared. 4x times positive 5y is combine our like terms, the xys and the xys, to get 16x squared plus 40xy plus

## A Different Pattern

Let's try and figure out one last pattern with our multiplication. We're going to start by multiplying each pair of these binomials. Notice that the first term is the same in each of the cases. Also notice that the second term is the same for each of the cases, except the signs are different, One's a plus, and one's a minus. Use what you know about multiplying binomials to find the product of each of these.

## A Different Pattern

If you got any of these correct, great work. We can use our area model to figure out the product of these two binomials. We know x time x is x squared, x times negative 4 is negative 4x. Positive 4 times x is positive 4x, and positive 4 times negative 4 is negative 16. We use a similar approach to figure out this multiplication problem and this one.

## The x terms

When looking at each of these three problems, what do you notice about the linear terms? What do the x terms add up in each of these cases? Write your answer here.

## The x terms

The linear terms add up to zero, if you got it, great work. We can see that for this problem the positive 4x and the negative 4x add up to 0. This middle terms are linear terms, sub is 0. So we're left with x squared minus 16. The same is true down here for this case and for this third case.

## Difference Pattern

So, using the patterns that you see here. What do you think a plus b times a minus b would be equal to? Think about what you're doing with the first term and think about what you're doing with the second term, to get your answers.

## Difference Pattern

Well, it turns out it's just a squared minues b squared. We know that these middle terms, or linear terms, always sum to zero. So we're really only left with these, our squares. So we'll have our first term squared minus our second term squared. That's true in every case.

## Difference of Squares

This pattern is called the difference of squares. Because here, we have two squares, and we have the difference of them. A subtraction sign or a minus sign means difference. Using our pattern, what's the product of 8 minus 3y times 8 plus 3y? Put your answer here.

## Difference of Squares

If you got 64 minus 9 y squared, excellent work. This problem was a little bit tricky, because I switched the signs on the parentheses. But remember with multiplication, we can switch the order of our parentheses, 2 times 3 is the same as 3 times 2. The same is true for more complicated expressions, so the 8 is my a, and the 3y is my b. A Squared would be 64, we just square 8. Then we have a subtraction sign and then we have b squared, which is 9y squared. Good thinking if you got it.

## Factored Form

Throughout this section, we moved from factored form to standard form. We multiplied binomials to find the product, or answer. The key idea here is that both of these forms are equal. They are a different way of writing the same thing. As factors, these represent the sides of a rectangle. But right now as a sum, this represents the area of the rectangles. I think this is an amazing part about math. It's two ways of writing the exact same thing. This idea of factors is really important, and it's one we're going to see throughout the rest of the course. In fact, in just the next unit, instead of going from factored form to standard form, we're going to go in the reverse direction.