[Music] Hi guys, in this video we continue the discussion on polynomials that we began in the previous videos in the playlist by talking about the so-called notable products and we will look in particular at the square of the binomial and the sum by difference. Let's start by saying that the term notable products refers to a series of simple rules that allow us to perform some common operations involving polynomials much more quickly and that can then be very useful for factoring polynomials, as we will see in detail in the following videos. Without wasting any more time, let 's start by looking at the first of these little rules and then talk about the square of the binomial. So let's assume we are asked to solve a + b squared. Now, in the absence of better ideas, that is, if one hasn't memorized the little rule we'll see shortly, what can one do? Of course, you may remember that squaring a certain quantity is equivalent to multiplying that quantity by itself. And at this point, by carrying out the multiplication we get a * a which is a^2, then we have an a * b which gives ab, a b * a which gives another ab and then we have a b * b which gives a b^2. And as you can see we have obtained two equal terms which we can at this point add obtaining a² + 2ab + b^2. Now, what are these three terms that we have obtained? Well, you see, a qu is nothing more than the square of the first of the terms we had in the parenthesis. Then we have twice AB, which would be double the product of the two terms we had in the parenthesis and then there is b², which is the square of the second term we had in the parenthesis and this result, or rather this little rule that we just discovered, remains valid for any pair of monomials I go to substitute in place of the ones I have called A and B here. And so if one memorizes it, one can fly directly from the square to the result without having to go through all these intermediate steps each time and therefore you understand with a considerable saving of time. To better consolidate our ideas and see how we can concretely use this result, let's immediately take a look at a couple of examples and start with the first one in which we are asked to calculate 2x + 3y, the whole squared. As you can see, we are in exactly this scenario, only that instead of having A and B we have this time 2x and 3y and therefore it will be enough to apply the rule we saw before by putting the monomial 2x in place of A and the monomial 3y in place of b . Proceeding we will then have the square of the first term, the double product of the first term by the second, so twice 2x * 3y and then here at the bottom there is the square of the second term and therefore by carrying out the two squares and doing the multiplication we obtain 4x² + 12xy + 9y². Of course, when one is trained, one jumps directly from here to here. I've now brought you this intermediate step where you can see I've highlighted the three terms we were talking about before in colors, but it's clear that these are all operations that one can then do mentally, thus jumping directly from here to here. Similarly, if we were asked to calculate 2x - 3y squared, we would be in this scenario again, only this time in place of a and b or, respectively, the monomial 2x and the monomial -3y. And so by analogy first, doing the calculations, we get the square of the first term, that is the square of 2x, the double product of the first term by the second, that is twice 2x * -3y. And here at the bottom we have the square of the second term that we had in parentheses. And if we now carry out the two squares and multiply here we get 4x² - 12xy + 9y². Naturally, the same argument applies as before: when someone is trained, when someone has understood, they jump directly from here to the result, applying the little rule and without any need to write down these intermediate steps, which I am providing here because I hope they clarify the flow of reasoning that leads to this result. I would also like to point out that the only difference between the square we solved before and the square we solved now is the sign of the second term in parentheses, and from the point of view of the result, the only thing that changes if you take a look is the sign of the double product, which was previously positive and has now become negative. So, in general when the two monomials that we put in place of A and B have the same sign, we should expect a double positive product. When the two monomials in place of A and B have opposite signs, we should expect a double negative product. As for the two squares, those will naturally always be positive, precisely because they are squares. Finally, let's look at one last example and suppose we are asked to calculate the square of the binomial -2x^ 2 + 1/2 xy. Again, we are in a scenario similar to this where instead of a and we have the monomial -2x^ 2 and the monomial 1/2 xy. And so reasoning as before, we will have the square of the first term that appears in parentheses, the double product between the first and second terms that we have inside the parentheses and then here at the bottom there is the square of the second term that we have in parentheses. And if at this point we do the calculations, we get 4x ^ 4 - 2x ^ 3 y + 1/4 x ^ 2 y ^ 2. Note that the double product came out negative precisely because the two monomials that we have in parentheses, as we were saying a moment ago, have opposite signs and naturally it doesn't matter if the first is negative and the second is positive. What matters in determining the sign of the double product is whether the two terms in parentheses have the same sign or not. But that said, before proceeding further, it is worth making a couple of important observations. The first, which follows immediately from what we were saying earlier, is that squaring a certain sum is different from summing the squares of the addends, and the reason, of course, lies in this double product which is not shown here. And the second thing I wanted to tell you is that if we assume that A and B represent positive quantities, then we can also give, if we want, a geometric interpretation to this result. And to do so, just take a look at the figure on the right. Imagine, in fact, that a + bresenti is the length of the side of a square. Okay? So this is a square with side a + b. Now, as you can see from the figure, we can break down our square into a small square with side A, a small square with side B and then there are two rectangles that have dimensions A and B. And so you see the area of my starting square , which of course in numerical terms would be a + b squared, I can think of it as the sum of the area of this small square which would be a qu, the area of this small square which would be b² and then the areas of the two rectangles, each of which would have area AB. And so here is the term 2ab that you see here. Given this, let's finally take a look at the second remarkable product, which is known as the sum by difference precisely because it is the sum of two terms multiplied by their difference. Now, if we carry out this multiplication term by term, we get an a * a which gives as a result a qu, then we have an a * - b which gives - a b * a which gives a + ab and then we have a b * - b which gives - b². And as you can see, of the four terms that appeared here, two simplify together and so we end up with an a² - b² as a result. And what are this a qu and this b²? Well, you see, these are the squares of the two terms that we originally had inside our parentheses, and so, what did we discover? That the product of the sum of two terms and their difference gives as a result the difference between the squares of the two terms. And if we now memorize this result as if it were a little rule, the next time we encounter a similar scenario we can jump directly from here to here without having to go through this intermediate step. To help us get a better idea, let's immediately look at a couple of examples and suppose we are asked to calculate 3x + 2y 3x - 2y. As you can see, we are in exactly this situation here, only that instead of having A and B we this time have 3x and 2y so in light of what we were saying a moment ago we know that the result of this operation will be the difference between the square of the first term, that is the square of 3x, and the square of the second term, that is the square of 2y. And so doing the calculations we conclude that the result is 9x² - 4y². Similarly, if you were to solve 2/3 x - 1 * 2/3 x + 1 and you see that we are in this scenario again, only this time we have the difference first and then the sum of the two terms, again the result will simply be the difference between the square of the first term, therefore the square of 2/3 x, and the square of the second term, therefore the square of 1. And by doing the two squares we conclude that the result is therefore 4x^ - 1. Naturally, in both the previous example and this example, you could have jumped directly to the result by skipping the intermediate step once you understood how things work because these are calculations that can easily be done in your head without needing to write them down. No, of course there is no problem if the first term between the two is negative. So, for example, if I were to solve -x³ + z * -x³ - z, reasoning similarly to the previous examples, I can say that this is the square of the first term, so the square of -x³ - the square of the second term, that is - the square of z. And so the result will be x ^ 6 - z ^ 2. It is also worth noting that the notable product works even if instead of A or B there is not a single term, but perhaps there is a sum of two terms. And to understand what I mean, let's take a look at this example here. You see that in the first parenthesis we have x + y to which 1 is added and in the second we have x + y to which 1 is subtracted. And so, you see, we are in exactly this type of scenario where our a is represented by the sum x + y, while our b is the number 1. It follows then that the result of this multiplication will be the square of our a, that is the square of x + y - the square of our b, that is - the square of 1. And so we conclude that the result is x² + 2xy + y², which would be this square here - 1. Before saying goodbye I wanted to take this opportunity to make a couple of interesting observations. But first, and I recommend you pay attention to this fact: if both terms inside the two parentheses you are multiplying change sign, then it is no longer a sum by difference. And so in a scenario like this where you see we have a minus b being multiplied by b minus a, the result is not the difference of the squares of the two terms as you get in this scenario here, but what you get, you see, is - the square of b minus a and the reason is that if we rewrite a minus b as the opposite of b, thus taking a minus outside the parenthesis, you see that there are two equal parentheses that give rise to this square. and not the difference of squares as happened before. So please pay attention to this fact here. While the other thing I wanted to tell you is that even this remarkable product can be, if we want, interpreted geometrically, always under the hypothesis that A and B are positive quantities. And above I have reported the corresponding graphic diagram. As you can see, a rectangle, having the long sides a + b and a - b respectively, which are exactly the two terms we are multiplying here, can be broken into two sub-rectangles, the green one and the yellow one, if we want, which we can then naturally rearrange in this way here. And as you can see from the two drawings, the sum of the areas of the two rectangles, or the area of the starting rectangle, which we know to be a + b * a - b, is equal to the area of a hypothetical square with side A. We must subtract the area of a hypothetical square with side B and it is with a² - b² that we find ourselves in the formula. That said, guys, I'm done for now . We'll continue the discussion in the next video where we'll talk about the cube of the binomial and the square of the trinomial. As always, if you find these lessons helpful, remember to like them, visit me on Facebook and Instagram, and check out my channel for tons of other videos. [Music]