 So now we're ready to talk about the so-called binomial theorem. The binomial theorem gets its name because we want to talk about how do you multiply out the expression x plus a to the nth power. X plus a is a binomial. If you want to take the power of a typical binomial, what happens? Well, it might seem weird to start here, but I'm going to start with the zero with power. If you raise something to the zero with power, you should get one. So x plus a to the zero with power is going to be one. Great. Then you look at the next one, x plus a to the first. Okay? Well, if you take something to the first power, there's really nothing to expand. You're just going to get x plus a right here. Things get a little bit more interesting. We get x plus a squared. x plus a squared, what does that mean here? Of course, that means you're going to take x plus a times x plus a, and you have to foil that thing out. You're going to get x times x, which is an x squared. x times a, and then an a times x, and then an a squared. When you combine like terms, you're going to get x squared plus 2ax plus a squared. All right? x plus a cubed. How does one do that? Well, x plus a cubed would mean x plus a times x plus a times x plus a. Now, the first one, x plus a squared, we already did that one. That's above. So we get x plus 2ax plus a squared. If you multiply those things out, you're going to get, distribute the x squared. You're going to get x cubed. Distribute the x squared onto the a. You get, I'm combining some like terms right here. Oh, I went the other way around. Sorry. We're going to distribute the x first. So x squared times x is x cubed. x times 2ax is a 2ax squared. x times a squared is going to be an a squared x. Then distribute the a. You're going to get an ax squared. Distribute the a on the 2. You get a 2a squared x. And then lastly, you get an a cubed, like so. If we combine like terms, there's x squared that can combine. There's x's that can combine. You're going to get an x squared plus 3ax plus 3ax squared plus a cubed. To do the next one, it's a little bit more complicated. What you would do is you would take x plus a cubed times x plus a. We multiply this thing out. We're going to take the coefficients from this guy right here, combine like terms, and you're going to end up with just the, I'm just going to list the final product. You're going to end up with x to the fourth, 4ax cubed plus 6a squared x squared, 4a cubed x and a to the fourth. I kind of hit the magic there, but I want to show you something that's going to be pretty neat. So if you look at the coefficients in the sequence, the first one was a one. When you take the zeroth power, when you take the first power, you get one and one. When you take the second power, you get one, two and one. When you take the third power, you get one, three, three and one. And although I didn't do the details for all these, if you do the fourth power, you can confirm it yourself. You're going to get four, six, six, four and one, which these numbers look very familiar. Aha! This is none other than just Pascal's triangle, right? One, one, one, two, one, one, three, three, one, one, four, six, four, one. When you start multiplying out these polynomial expressions, these binomial powers, the coefficients you get are just going to be the numbers in Pascal's triangles with those numbers are called the binomial coefficients. They're called the binomial coefficients because of the binomial theorem. So when one multiplies out something like x plus a to the nth power, what are we expecting? Well, we're going to get a bunch of products that look like the following a to the j times x to the k with the property that j plus k equals n. So we look at all the possible combinations. So like look here, we have x to the fourth, we have one a three x's, two a's, two x's, three a's, one x, zero x's, four a's. So we look at all the possible ways of combining two numbers, two natural numbers to make four. And you see there's five possibilities, zero plus four, one plus three, two plus two, three plus one and four plus zero. Those are the two possible ways. So that's what these monomials are going to look like in this expansion. Then the coefficients are just the nth row of Pascal's triangle using the binomial coefficients. So this is then the binomial theorem, x plus a to the nth power. It'll look like this polynomial where the powers of x's are getting smaller. The powers of a are getting bigger until eventually we end up with, why is that an a? That should be a two, a squared. And so in the end, you go from a to the zero up to a to the n. And then the coefficients are just the binomial coefficients from Pascal's triangle. It's a beautiful little theorem. Let me put it into practice right here. We're going to use, oh boy, typos galore here, we're going to use the binomial theorem to expand x plus two to the fifth power. So thinking of Pascal's triangle, I'm going to need that. So we get one, one, one, two, one, two, one, one, three, three, one, one, four, six, four, one. And then the fifth row is going to look like one five, 10, 10, five, one. Of course, if you have Pascal's triangle written in front of you, you don't have to go through these calculations over and over again. So by Pascal's, by the binomial theorem, what we get is the following. x plus two to the fifth is going to look like five choose zero times two to the zero x to the fifth. Then we're going to get five choose one times two to the first x to the fourth plus five choose two x are two squared x cubed. Then we're going to get five choose three times two cubed x squared. Then next we're going to get five choose four, two to the fourth x. And then finally we get five choose five, which is going to be two to the fifth x to the zero. So simplify these things five choose zero from Pascal's triangles of one two to the zero is one, and we get x to the fifth. The next one five choose one is a five two to the first is two, we get x to the fourth. Next one five choose two is 10 times two to the fourth, which is four times x cubed. And the next one five choose three is also 10. Then we get two times two cubed, which is eight times x to the squared there. Then we're going to five choose four, which is five times two to the fourth, which is 16 times x. And then finally we get five choose five, which is one times two to the fifth, which is 32 times x to the zero. So you don't see any x is there. And then simplifying the coefficients, we're going to get x to the fifth, five times two is 10, 10x to the fourth, we get 10 times four, which is 40 x cubed. Next we're going to get 10 times eight, which is 80 x squared. The next one five times 16 notice if you borrow one of the 16s to you get two times five, which is 10. So you get eight times 10, which is 80, 80x. And then lastly you're going to get 32. So when you compare that to the alternative, that was a whole lot easier to multiply it out that binomial expression x plus two to the fifth. It was super simple. We didn't have to foil anything at all because we solved this pattern here. Let's do one more example. I have it hitting down here. So let's expand two y minus three to the fourth right here. And so the expansion is going to look something like the following. We're going to get four, choose zero times two times y to the fourth power times negative three to the zero power. Don't forget the negative sign. Then the next, we're going to get four, choose one times two y to the third times negative three to the first. Then we get four, choose two times two y squared negative three squared. Next, we're going to get four, choose one, excuse me, four, choose three, which admit it was the same number, but let's go in the right order here. You're going to get two y to the first, and then you're going to negative three cubed. And then lastly you're going to get four, choose four, which is going to look like two y to the zero, then negative three to the fourth, like so. So for the binomial coefficients, we can compute them using factorials, or we can use these Pascal's triangle that's right here. The fourth row looks like one, four, six, six, one, so we'll just remember those numbers. And so we end up with one times two to the fourth is 16, 16th y to the fourth, and then we times that by negative three to the zero, that's one. Then the next one, we're going to, binomial coefficient was a four, we're going to get two cubed, which is eight y cubed times negative three to the first, which is negative three. Then the next one we're going to get is four, choose six, or four, choose two, which is six. Two y squared will give us a four y squared, like that. And then you get negative three squared, which gives a positive nine, pay attention to the sign there. The next one, four, choose three, which is also four. Then we're going to get a two y, and we're going to get negative three cubed, so that's negative 27. And then lastly four, choose four is one, two y to the zero is also one, and we get negative three to the fourth, which is a positive 81. And so multiplying, there's no like terms to combine here, that's already happening. So multiplying, we're going to get 16 y to the fourth. Next we're going to get minus, let's see what we have here. We're going to have eight times three, which is 24. If we times that by four, that's just doubling it twice. So you're going to get 48 and then 96, so 96 y cubed. The next one, we're going to get positive four times six times nine y squared. And so we get four times, excuse me, four times nine, that's going to be 36 times that by six, you're going to get six cubed, which is 216 y squared. The next one's going to be negative. You have four times two times negative 27 y there. If you double 27, you get 54. Then two times that by four, we have to double it again, right? So that's going to be 108 and then 216. So we get another 216, negative 216 y. And then lastly, one times one times 81, that's an 81. So a little bit of arithmetic that has to be done. But for the most part, this is fairly painless compared to what we'd have to do alternatively, alternatively there. The binomial theorem is a super useful tool to simplify multiplication of binomials raised to various powers.