Multiplying Polynomials - Video Tutorials & Practice Problems

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1

concept

FOIL

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3m

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Welcome back, everyone. We already saw how to add subtract polynomials. And in this video, we're going to start talking about multiplying polynomials. It turns out we've already done some of this stuff before. When we looked at the distributive property with expressions and exponents, we saw how to take terms on the outside of parentheses and distribute either constants or variables on the inside. But a lot of problems are not going to look like that. A lot of problems are actually gonna start looking like this where you have two parentheses that are multiplying together. So a lot of problems are actually gonna involve multiplying two binomials. Remember these are just expressions with two terms. And so when, when we have these kinds of problems, we're not gonna use the distributor property, we're gonna use a new method called the foil method. And that's a term you may have heard in a math class somewhere. We're gonna go over it and I'm gonna show you that it's actually a very straightforward procedure that tells you exactly what to do and the order. All right. So let's just go ahead and jump right in foil is actually just an acronym and it tells you which two terms to multiply with this expression over here, these two, these two binomials and it also tells you what order to multiply them in. All right. So the foil, they stand for different things, let's just jump right into it. The f actually stands for the first terms. So you multiply the first two terms of each expression. In other words, that would be just the X and the X. The next thing you would do is the o which is the outer terms. So if you think about this, there's sort of four terms in this expression or this uh multiplication, the outer terms are the two terms sort of on the outside. So the X and the three, all right. Now, the third thing is you're gonna multiply the I, which is the inner terms that would actually just be the two and the X that's on the inside right here. And the very last thing is you multiply the last terms. So you just multiply the two and the three notice how every single term uh gets multiplied by the two terms in the opposite expression. So everything actually gets multiplied by, by each other. Um And that's how you sort of do it. So let's just go ahead and, and foil this expression out. How do you multiply X plus two and X plus three? Well, remember F means that we're gonna multiply the first two terms, the X and the X, what happens when you get that, you just get X squared. So let's do the o which is the outer terms uh that would just be the X and the three well, X and three multiply together just becomes three, X. What about the inner terms? The two and the X, that's the I, so the two and the X just becomes two X and finally, the two and the three, those are the last terms, two and three just becomes six, not six X because there are just two numbers together. All right. So that's it. That's the whole thing. That's how to foil. The very last thing you have to do is once you foil, you usually gonna have to simplify your expression because usually what happens is these middle terms will be like terms and you'll be able to combine them. All right. So in other words, this expression simplifies down to X squared. I can't combine that with anything else, but I can combine the three X and the two X because they're like terms and they actually just combine down to five X. And then finally, you just have the six that comes down like this and that is your simplified expression. So this is how you foil two binomials and simplify it. Let me know if you have any questions. If not, let's move on to the next video

2

Problem

Problem

Multiply the polynomials using FOIL.

$\left(x-5\right)\left(x-12\right)$

A

$x^2-12x+60$

B

$x^2-17x+60$

C

$x^2-5x+60$

D

$x^2-17x-60$

3

Problem

Problem

Multiply the polynomials using FOIL.

$\left(4x+7\right)\left(-x+6\right)$

A

$4x^2+17x+42$

B

$4x^2+31x+42$

C

$-4x^2+17x+42$

D

$-4x^2-x+42$

4

Problem

Problem

Multiply the polynomials using FOIL.

$\left(x^2-3x\right)\left(2x+8\right)$

A

$2x^2+2x-24$

B

$2x^3+2x^2-6x-24$

C

$2x^3-2x^2-24x$

D

$2x^3+2x^2-24x$

5

concept

Multiply Polynomials Using the Distributive Property

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4m

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Welcome back everyone in this video. I'm going to give you a summary of all the different ways that we multiplied polynomials. For example, we saw how to distribute one term that's outside of parentheses by making sure that it multiplies into everything on the inside. We also saw in the last few videos how to foil something. That's where you have a two term times a two term expression, but some expressions aren't gonna look like that. In fact, sometimes you might have many terms like this two term expression multiplied by a three term expression. And I'm gonna show you how to work through this. And basically, we're gonna see that it's very familiar. Let's check it out. So why can't I take an expression like this and foil it? Well, if you try to do this, what's gonna happen is you'll do first and you'll do outer and then you'll do inner and then last, the problem is that once you've multiplied everything, this term over here on the inside, actually never got multiplied by everything you've kind of left it out. So foiling only actually works very specifically for two term times, two term expressions. So that's the whole thing. Can't foil here. If you try to foil it, you're gonna get the wrong answer. So, how do I deal with something like this? Well, basically the idea is that when you multiply polynomials with more than two terms, all you really are gonna do is you're gonna split the terms of the shortest expression and you're gonna turn it into a distributive property type problem. So, uh sorry, then distribute. So here's what I mean by this, we're gonna take this smaller expression like X plus three. And I'm basically just gonna break it up into two things. I'm gonna have an X over here and a three over here. And I'm basically just gonna break it up into two distributive property type problems. So what goes on the inside here? It's this whole entire expression X squared plus X minus two and then X squared plus X minus two. All right. So basically, what I'm doing here is I'm making sure that this X gets multiplied by everything that's on the inside. That's what happens in this first expression. And then the second one making sure that this three multiplies everything that's on the inside and that's what the second expression does. All right. So basically, what we've done here is we've turned uh this problem back into just one of the distributive property problems. That's the whole thing. The rest really is just doing the multiplication and stuff like that. So let's actually go ahead and do it. Well, actually, for this one, I don't have to because notice how this is the exact same expression that I have over here. So this is what that becomes. I'm just gonna rewrite this here. X cubed plus X squared minus two X. And what about this three? Now we just have to distribute the three into the X squared that becomes X three X squared, then the three into the X becomes three X and then the three into the negative two becomes negative six or minus six. All right. So now the rest is really just simplifying. But um let's go ahead and do that. So I see an X squared term over here but then I see an X or sorry X cubed and I have an X squared with an X squared and I see a two negative two X and a three X and then I see a negative six off by itself. So remember I'm just gonna combine like terms I have to combine things that are similar. And what this becomes here is this becomes X cubed. Uh And then the X and the three X squared becomes uh four X squared, the negative two X and three X become just X and then I have negative six over here. So this is my simplified expression and that's it. That's how you multiply polynomials just in general. And that's really all there is to it. So I'm actually just gonna point out to, uh, point out a pattern that's happened here, which has to do with the number of terms that we initially started with and ended with. So in other words, when I took something like this, like a term outside and multiplied it into a three term expression, I did one times three. And what I ended up getting here was I end up getting three terms, uh, uh, as my answer. And when I foiled, I did something similar. So I took a two term and a two term problem. And when I did the multiplication out, I ended up with four terms over here before I simplified it. The exact same thing happened over here. I took a two term expression multiplied by a three term expression and I end up just getting six terms before I did the simplification. So it's a really good way to check really easily when you've done the multiplication correctly. Always just take the number of terms that you have multiply them and whatever number you get. That's how many terms that you should get uh before you start simplifying everything down. If you don't have that, that means you dropped some, you missed one or you may have counted an extra one.

Welcome back everyone. So by now, we've seen how to multiply polynomials by either foiling or distributing. What I'm gonna show you in this video is that some multiplication problems will actually fit a special pattern. And when this happens, we can use formulas, uh these things called special product formulas to multiply because it's gonna be way easier and way faster than having to do stuff. The long and tedious way by foiling or distributing, I'm gonna show you that you can take a look at a problem like this and very quickly you'll be able to tell me that the answer is X squared minus 25. I'm gonna show you how it works. So basically, if I take a look at an expression like this, the only way I know I know how to multiply it right now is by foiling, that's what I showed you. That's the uh first and outer and then do the inner and last. So I'm gonna multiply all these terms out to show you. Uh this just becomes X squared minus three X plus three X plus and then this actually minus nine. All right. And if you simplify this, what you'll see is that the negative three X and positive three X cancel and all you end up with is X squared minus nine. So that was a lot of work just to figure out that two of the middle terms will cancel. And the answer is relatively straightforward. It's just two terms. Now, what actually happens is the reason this happened is because these two terms are the same but just opposite sides. And whenever that happens, the middle terms are always actually going to cancel. So this is like a little bit of a pattern. This always happens as long as these two numbers are the same. And so we've come up with actually a special formula for this. And it's basically if you can sort of see if you can detect that a multiplication problem is fits fits the sort of pattern of A plus B times A minus B. The answer will always just be A squared minus B squared. This equation pops up a lot in math. We actually give it a name. It's called the difference of squares, but you don't need to know the name. So basically the whole thing is you could take a look at a problem and if you can figure out that it fits this pattern, then you could just go straight to this rather than having to foil it. And that's gonna be your answer. So let's take a look here. I have X plus five times X minus five. So in other words, I have the same number, just opposite signs. So this is like A plus B A minus B. So the answer is, is just gonna be a squared minus B squared. So basically, what happened is that like my, is that A is my X and B is my five. And so therefore, what happens is when I take A squared that just becomes X squared. So these two things are the same and when I do B squared, that just becomes 25. All right. And this is just your answer. All right. So basically the way that these problems work out is it's kind of like pattern matching. Um anytime you see a multiplication problem and you see the same numbers but opposite signs or you see stuff that's squared, usually it's gonna fit one of these formulas and you could just go straight here instead of having to multiply everything out the long way. Let's take a look at this next problem here. X plus six squared. All right. This actually fits a special pattern uh of A plus B squared or A minus B squared. These equations are called the squares of binomials. And I'm just gonna show you how these work out. Um This is basically what the right side equals and we have a plus and plus as the two signs and a minus and a plus uh one easy way to remember this, by the way, is that the first sign will always be the same as the left side. Um And the second sign will always be positive. So that's one way you can remember this real quick. All right. So if I take X plus six squared, which one is this, which, which one of the two patterns is this? Well, it's clearly this one over here. So I'm just gonna basically use uh my formula. So this is A and this is B. So in other words, my A is xib is six. So what the answer is supposed to be is the answer supposed to be A squared plus two A B plus B squared. All I have to do is now just, you know, replace these letters here with um with their XS and sixes and stuff like that. So my A squared just becomes X squared. Uh What about this? Two A B? Well, this just becomes two. Then what's A A is S A is X and then what's B uh B is six? And then plus B squared, B squared uh ends up being six squared, which is just 36. So if you simplify this, what you end up getting is X squared plus 12 X plus 36. And that is how to multiply this again, very quickly without having to foil this. All right folks. So that really, that's all there is to it.

9

Problem

Problem

Multiply the polynomials using special product formulas.

$\left(5x-9\right)\left(5x+9\right)$

A

$5x^2-81$

B

$25x^2-81$

C

$25x^2+90x+81$

D

$25x^2-90x+81$

10

Problem

Problem

Multiply the polynomials using special product formulas.

$\left(3x+5\right)\left(3x+5\right)$

A

$9x^2+15x+25$

B

$9x^2+25$

C

$9x^2+30x+25$

D

$9x^2+55$

11

concept

Special Products - Cube Formulas

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3m

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Welcome back, everyone in this video, we're gonna continue talking about special product formulas. Um And in this video, we're gonna cover some formulas involving cubes of polynomials rather than just squares. But the idea is the same, we're just gonna be using these special products rather than having to multiply out. The long way, let's check it out. So basically, I'm just gonna jump right into the formulas over here. We have uh cubes of binomials, which is A plus B cubed or A minus B cubed. Remember we talked about squares of binomials and now we just have cubes of binomials and, and there's just more terms, but basically the way that this works is that the signs go plus plus plus and then for a negative, you're gonna do negative plus and negative. So one way to remember this is that for plus, everything is positive and for minus, basically the signs alternate, it goes negative positive and the negative. OK. So, all right. So if I take a look at my example here, I have X minus three cubed. So which one is it? It's basically just the A minus B cubed. That's sort of the pattern that it fits. So in other words, this is like my A and that's my B. So in other words, X is A A and three is B. So what does this work out to? Well, this works out too. I'm just gonna write this out. A cubed minus three A squared, B plus three A B squared minus B cubed. The rest is just I have to figure out which each one of these things becomes now that I know that A is X and B is three. So let's get started. What is a cubed? Well, a cube just becomes X cubed. Now, what's the middle terms? So what is the first or the second term become, this becomes three? And then A squared, if A is X, then A squared is X squared and then B is just three. And then what about the third term? This becomes three A is X and then B is three. So B squared just becomes nine and then uh minus B squared. So that's three times three times three, which is just 27. All right. So notice how fast this was. If we had to foil it out, it would have been a nightmare because we would ended up with a three term. And you got to multiply that again, it just would have been a disaster. So these formulas are again, relatively speaking, pretty, pretty quick and straightforward. All right. So the last thing we just have to do is now simplify this. So the X cubed now just go straight down. And what does the second term become? I have a three times a three times an X squared? So this just becomes nine X squared. The middle, the third term actually goes to three times 27 times just one power X. So this is 27 X and then I have minus 27 over here. And that is the simplest I can make this, I can't combine any more terms or anything like that. All right. Um All right. So this is my expression over here. And that's how you do the cubes of binomials. Now with these types of formulas almost always or you're almost never gonna have to memorize these, your professors may actually give you something like a formula sheet, but in case they don't, I'm gonna give you a good way to think about this. Um And sort of memorize them, we've already talked about the signs. But if you actually notice what happens to the coefficients, the coefficients always go 133 and one, right. So I have a, one, a cubes of three and then three and then a one B cubed over here. So that, that's how the coefficients always go. There's also a pattern with the powers of A and the powers of B, the powers of A decrease as you go from left to right. So we have an A cubed then an A squared and an A to the one power. And that if you have no A over here, it's basically like you have a, the zero power. So in other words, the powers of A always go from 3 to 1 to 0. And for B, it actually goes the opposite. The powers of B actually increase. Notice how here we have A B to the zero power, then B to the one, then B to two and B to the three. So this goes 012 and three. So basically A goes down and B goes up and the coefficients are always 133 and one.

12

Problem

Problem

Multiply the polynomials using special product formulas.