College Algebra: Multiplying Big Products

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Okay, well, let's talk about some more stupid polynomial tricks. Now, in particular, you can factor two binomials using the foil method or whatever. But what about if you are given two big mama things, like for example, like binomial or trinomial or a trinomial and a gigol-nomial, you know, I mean how do you do that? Well there's no nice little way of spelling out the method. So instead of worrying about the whole foil thing what we use is the distributive property.
So let me actually show you an example of this. You know, actually, I sort of enjoy these things. That'll show you how deranged I am, because it is sort of fun. It's sort of bookkeeping. Now, of course, if you have something simple like this (-x + 2) (1 + x) you could just use the foil method. And if you do the foil method, I'll do it for you really fast, I'll just talk through the foil method. I see a -x + 2 - x^2 + 2x, and you combine that and you get the answer. Or, what you could do is you could use the distributive thing. And the distributive thing is to consider this whole thing. So if we take some of this stuff and think of this like a blob then we can just distribute that whole thing times 1 and distribute that whole thing times x. If I distribute anything times 1, it is just itself, so in fact, this would just give us -x + 2, that's what I get when I take the whole blob and multiply it by 1. If I now take the whole blob and multiply it by the x that actually required me another distribution. Do you see where it would be? It would be this 1. It would be x times all that. And so that would be -x^2 + 2x, distributing that across. So when you are foiling you are just taking every term here and multiplying it by every term here.
Now you take that method and you can now do much more exotic things. For example, let's multiply (y + 3) (y^2 - y + 1). This looks frightening at first, but just remember, think about this thing as a blob and just distribute that through to each term. Let's see what happens. If I distribute the blob to here I would have, blob times y^2, then -blob times y, then +blob times 1, I'm going to suppress the desire of writing 1 in there. So, I've got the blob here, the blob here, the blob here, let me write the blob in that's y + 3, y + 3, and y + 3. Well, now I do the distributive property for the other way, you see. I've got now this thing as sort of the thing, and I'm going to distribute this way, so this would actually be y^3 + 3y^2. Now here what do we have? Here we have a y^2, if I distribute, I'd see a y^2 and then I'd see a +3y and don't forget we are subtracting so I'm going to put a minus sign there, and then I've got that last step there.
Now, does this look right to you? Let me help you, it's not right. In fact I made one of my favorite mistakes of all time. Do you know what that mistake is, it's number four on my top ten list, that's right, number four--subtracting mistake. Classic mistake and it's number four on my list out of 10, and that is forgetting to distribute the negative sign. Remember the moral, there it is, spread the negativity, you have to multiply everything through by that negative sign you even have to subtract that. And now what do we have? We have y^3 + 3y^2 - y^2, I'm spreading that now with distributive, -3y + y + 3, such a big problem I'm running off the page. And what do I see? I see y^3 and I'm just combining like terms. What squares do we have here? We have a 3y^2, we have a -y^2 so that's 2y^2 and that's it. So then + 2y^2 any y's in the picture? Yep, we have a -3y and then we have a +y, so that's a -2y and then we have a +3. And that's the answer so, okay, it is a little bit messy I admit, but you cover this up, cover that up, that's what multiplying all this out equals.
Notice all I'm doing is using the distributive property. Now you could do much more exotic examples, in fact, maybe I'll try one really, really, really big one. How about x - y + z, so that's a trinomial and I'm going to multiply it by another trinomial 2x^2 - 4y + z, so two trinomials together. One way of thinking about this is to say, you know what, every term here has to be multiplied by every term here, or, you could think about it as the blob, this blob times that, this blob times that, this blob times that. But notice that if you are taking this blob times this, then later on in life I'll have to take this times each of these things. So, actually, if you don't like the blob or you don't have this kind of putty around to use this, you could just be really, really careful with the bookkeeping and make sure every term hits every other term.
Let me show you this way, it's the same way, but it's a different way of thinking about it and also, it takes up a lot less room. This x right here has to multiply every single term here, so let me just do that, x times this is going to be 2x^3, x times this is -4xy, x times this is +xz, there I've taken that x and pushed it all the way through to these terms. Now let me take the -y, notice there's a minus sign there, the -y and hit it through every term here, so -y times this would be -2x^2y. Then -y times this would be a plus, because a minus times a minus is a positive, +4y^2 and then a -y times these is going to be -yz. So there's the next whole terms that we have to all add together. So now I have pushed the -y all the way through and I've got that z left so z has to hit everybody. And so I'd see +2x^2z - 4yz + z^2 so all those terms added together is what this works out to be.
By the way, one little closing thing here. How many terms are there? How many monomials are there? One, two, three, four, five, six, seven, eight, nine. And in fact if you think about it a little bit you could always make sure that your answer is right, at least when you have it out in its expanded way, because since one of these three terms has to multiply each of these three terms there should be a total of what? Well, let's see. There are three things here, and three things here, and everyone has to hit each other so there's actually nine possibilities. So in fact, if we only had eight or six terms we know that we did something wrong. And of course after you have the terms you might be able to combine them. And in that case they might actually get less, so don't panic if the final answer, after you simplify everything, is less. But when you multiply it all out you should check and say, okay, look, there's three things here, three things here, I put it all together I should get nine.
For example, with two binomials using the foil method how many should you have? You should have four terms, right, two and two, and that's what you have. Foil, spell it--four. Okay, see you soon.
Prerequisites
Polynomial Expressions
Multiplying Big Products Page [1 of 2]