Common Core Algebra I.Unit 1.Lesson 9.More Structure Work.By eMathInstruction

Hello and welcome to another common core algebra one lesson by E math instruction. My name is Kirk weiler, and today we'll be doing unit one lesson number 9 more structure work. Before we begin, let me remind you that the worksheet that goes with this lesson. And a homework set can be found by clicking on the description of the video. As well, don't forget that on our worksheets we have a QR code that can be scanned with either a smartphone or with a tablet to bring you right to the videos. All right, let's begin. One of the most important things in common core algebra one is seeing structure and expressions. In other words, being able to look at parts of an expression and treat them as a single entity. So what we're going to do is we're going to do this work like we did in a previous lesson, but it's going to be a little bit more complex still. I want you to try to have fun with it because I think it's kind of cool. Let's take a look at the first exercise. It says consider the somewhat complex expression. X times X plus four plus two times X plus four. Letter a says write an equivalent trinomial expression. Test the equivalency with a value of X equals one, show the test. All right, well, one thing that we could do immediately. And I'm going to rewrite the actual expression down here. So that we have something to work on. Is we could distribute. We could say, well, I'm going to take that X and on a multiply it through here, and I'll get X times X, which is X squared. And then X times four, which is four X here, I could distribute this two. So I'll get two times X, which is two X and two times four, which is 8. Then I can combine these two terms, and I can get X squared plus 6 X plus 8. So I've done not double distribution. I've distributed twice so maybe it's double distribution, and I've gotten this particular expression. Now in theory, this trinomial one, two, three terms, is equivalent to this more complicated expression. And here comes my test. All right? I'm going to test it with X equals one, because I've been told to test it with X equals one. And first thing I'm going to do is I'm going to substitute it into our sort of more complicated expression. X times X plus four. Plus two times X plus four. So what do I get when I evaluate this expression with X equals one? Well, I get one. Times one plus four. Plus two. Times one plus four. So that'll be one times 5 plus two times 5. So that'll be 5 plus ten. That will be 15. Now let's see what I get when I substitute one into this trinomial expression. Well, here I'll get one squared plus 6 times one. Plus a one squared is one times one. 6 times one is 6. 8 is just a and one plus 6 7. Plus 8 gives me 15. Awesome. Now, by the way, that test doesn't prove that these two expressions are equivalent. We could have been some of the most unlucky people on earth and gotten the wrong trinomial, but it just happens to have the same value as our original expression when X is one. One thing I know for certain though is that those two don't turn out to be the same, then I've done one of two things. I've either botched it when I did the test. Or I botched it when I came up with the equivalent expression. Unfortunately, I'm not really sure which. Let's take a look at letter B writing equivalent expression that is in the form of the product of two binomials, all some tests the equivalency with X equals one. Well, this is really cool. Plus there's this random line that just appeared on the screen. I don't know where that thing came from. And not really sure why I'm even trying to erase it. Okay, just decided that's not a great move. Here's what I want to bring your attention to because we're going to see this a lot today. Notice that there's an X plus four and an X plus four here and here. So what I'm going to do is I'm going to actually reverse the distributive property, and I'm going to factor that X plus four out. Now what was that X plus four multiplying? Well, it was multiplying an X that's here. It's kind of hard to see. Through that red box in there, right? So it was multiplying the next. And it was multiply a positive two. So there it is. I claim that this complicated expression is the same as X plus four times X plus two. I think I'm going to need this space signing to get rid of my line. All right, and again, we're going to test it. So we've already tested the expression for X equals one with the more complex expression. Let's now test it here. All right, so I'm going to put one into here, one plus four, and I'm going to put one into here. One plus two, of course one plus four is 5. One plus two is three. And that's 15. Yes. So what are we really doing here? What we're doing is that we're seeing that X plus four in letter B can be considered a single entity. And it really is because it's grouped inside a parentheses like that. We're treating it as if it's just one thing. That one thing is multiplying both an X and a positive two. We can reverse that multiplication by factoring an X plus four outward factoring here. All right, we're rewriting, rewriting this more complex expression into a pure product. X plus four times X plus two. And we're going to work with that quite a bit today. So I'll point it out each and every time. Anyhow, I'm going to clear this out, think about it a little bit, write down anything you need to, pause the video. All right, scrubbing. And we're scrubbed. All right, let's take a look at number two. Consider the expression X plus four times X -5. Plus X plus four times X minus two. Writing equivalent expression that is in the form of the product of two binomials, product of two binomials. Test the equivalency with a value of X show your test. Now, I bet that some of you have already caught on. So what I'd like you to do if you think you understand what's going on is pause the video and take as much time as you need to to try to answer this question. All right, let's go through it. So again, let's highlight what we should be seeing here. All right, what I notice is that the X plus four here is the same as the X plus four here. So just like in the last problem, I'm not sure why I like the X plus four so much. I'm going to factor an X plus four out. Now what's it multiply? Well, technically, the X plus four is multiplying an X -5. I'm going to leave it in parentheses for right now. And it's also multiplying an X minus two. Now we have most certainly not done quite yet with the problem is asking me for. This is definitely a binomial. All right. But this is not yet a simplified binomial. We want to kind of keep going a little bit on this. So I'm going to take that X plus four and I'm just going to leave it. Now what am I going to do here? Well, at first, when we look at this, those parentheses seem kind of confusing. In fact, we can use the associative property of addition and subtraction to really remove the parentheses. Say well, now I don't have to do it in that order. In fact, I can look at that problem the same way as I look at this problem. And then I can really look at this as X plus X I'm going to change this into plus negative 5 plus negative two. Some of you won't need to do this step. Some of you will be able to go to the final step almost immediately. But I'll get X plus four times two X -7. And there is the product of two binomials. Now, how do we know whether or not we've done it correctly? Well, we could grab a value of X, let's not go with X equals one. That's getting a little bit boring. Let's go with X equals, let's say 6. I like that. So we're going to test. All right. So let me test it first in the more complicated expression. The X plus four times X -5 plus X plus four times X minus two. Let's test it in this expression. So 6 plus four times 6 -5. I don't want to lose my parentheses there. 6 plus four times 6 minus two. All right, 6 plus four is ten. 6 -5 is one. Again, 6 plus four is ten. And 6 minus two is four, ten times one is ten. Ten times two four is 40. So that expression, the more complex one is 50 when I test X equals 6. Let's now test it in X plus four. Times two X -7. Well, if we put X equals 6 in there, we get 6 plus four. We get two times 6 -7. So this is going to be a little bit more complicated. Two times 6 is 12. -7. 12 -7 is 5. And ten times 5 is 50. And there we have it. Yes. Right? You can always test equivalency or more properly. You can always test whether equivalency is incorrect. That doesn't really prove that the equivalency is correct, but it does tell us that does tell us that we don't know whether it's correct or incorrect. I'm not sure how athletic. It's a good indicator that we've probably done it right. Okay? So I'm going to clear out the text. Here we go. All right, let's keep playing with this. This is really a form of factor, right? Where we look at these common binomials. And we factor them out. Pull them out using the distributive property. So take a look at number three. Now this one's going to be a little bit trickier. All right? I'll explain why it says which of the following is equivalent to the expression blue, show the manipulations that lead your choice. We have to figure out which one's equivalent to this. All right, so let's look at what's common here. We've got X minus three. And X minus three. So I'm going to factor that X minus three out. Okay? And what I'm going to be left with I'll just stick with red here is two X plus 7 minus X minus four. All right, let's talk a little bit about this. Okay, nothing that we have to worry about here. We can just write that down as two X plus 7. Get rid of those parentheses they're doing nothing. All right. But here, this is tricky. Here we are subtracting X minus four. So what we're going to do is we're really going to look at this as multiplying this entire quantity by negative one. Essentially, whenever we subtract a binomial or a trinomial, we have to switch the signs on absolutely everything. And that's really tricky. I have pre calculus students who mess that up. We can't just subtract the X, we also have to subtract the negative four. And when we subtract a negative four, it becomes a positive for. Now, doing a little bit of rearranging, we can say that that's two X minus X plus 7 plus four, and that'll be X minus three times X plus 11. Oh, there it is. All right. Again, we can test the equivalency by picking a good value of X and seeing if it works out correctly. All right? Watch out. I can't emphasize enough how tricky subtraction is going to be. We'll see that again and again. So we'll get some practice on it. Okay? So pause the video now if you need to. All right. Moving on. All right. Rewrite each of the following expressions as an equivalent product of two binomials. Wow. Okay. Here we go. So what I'd like you to do, if you think you understand how this works, is that like you to pause the video. Try to work through all 5 of these. I'll give you another chance to pause the video eventually. But it's good if you take a shot at first. Watch out for anything involving subtraction, especially letter B and letter E and especially especially, especially, letter E, letter E can be very, very tricky. All right? So give it a shot, pause the video. I don't know, I would say take up to ten minutes on this problem, at least, and see what you can get. All right, let's go through it. So the first few are actually quite easy. We just look and we say all right, I've got an X plus 5 and an X plus 5. So that's what I'm going to be pulling out. All right, in fact, we're going to be done with the first problem before we even know it. What's multiplying the first X plus 5 is X, what's multiplying the second X plus 5 is positive 7. And that's it. Each of these is a binomial. Remember, binomial means two terms. One, two, one, two. That's it. I think I'm going to switch back to blue. All right? Let's take a look at this one. X minus two, and X minus two. Right? What are they multiplying? Well, in the first case, it's multiplying three X and in the second case, it's multiplying negative four. And yet again, we're done. And just keep switching colors. Now letter C is actually quite tricky. Um. What am I supposed to do here? I mean, I see an X plus four here. And I see an X plus four here. But I don't see anything multiplying this. So one thing that we could do is we could rewrite this in kind of a tricky way. We could rewrite it like this. Now I want you to think about this for a second. Let's sit back and really appreciate what we've just done. You can multiply by the number one anytime you want. It's actually got a special name, not one that you particularly know, need to know. It's called the identity element for multiplication. Because when we multiply by the number one, it doesn't change anything. So I can rewrite X plus four as one times X plus four. The reason that's an advantage is that I can now say all right. I am going to factor out an X plus four, and then what I'm left with is a negative two X on the first term and a positive one on the second. And there it is. All right. Why don't we go back to red? Okay, for letter D, we look, we look, we look at an X plus three and the next plus three. Here's where it's going to take us a little more time. So we take that X plus three. We factor it out. We have an X -6. Actually, let me be good about this. Sometimes I have the urge to skip steps, and I shouldn't. Because really, technically speaking, I'm multiplying that X plus three by X -6, and by an X plus 9. Now, because it's pretty much all addition going on inside of here and by addition, I mean that one. I can actually remove the parentheses, X -6 plus X plus 9. When I write this out all the way, especially so we can contrast it with the next problem. So I'll get X plus X plus negative 6 plus positive 9. So I'm using the commutative and associative properties of addition there, plus the fact that I can rewrite that negative 6 as positive, negative 6, or plus negative 6. Sorry. And now I can finish this problem off X plus three. Times two X plus three, that plus three comes from combining these two. And I seem to have left off a parentheses. There it is. Remember in each case what we're doing is we're writing an equivalent expression and equivalent equivalent equivalent equivalent expression. All right, last one, love me some X minus four. All right, we factor that X minus four out, and the first case we're left with two X plus one. In the second case, we're left with minus X plus 6. Again, very critical. We can now look at subtracting that X plus 6. Nothing going on here. Absolutely nothing. But we distribute that multiplication by negative one. Some teachers will say that we're distributing the subtraction. I think that that's okay to say. I think that that's a good way to put it. And then we can look at it as two X plus negative X plus one plus negative 6. And two X minus X is X one -6 is negative 5. And we get X minus four times X -5. Oh, good. Still correct. And multi colored. I should use multiple colors more often. It looks good. All right. So that is tricky. Waiting for the next problem though. I think you're going to like this one. So let me clear out the text, pause the video now if you need to. All right, here we go. All right, last problem. I love this one. So this takes us back to something that we did in the lesson a few lessons ago. That doesn't sound right. But anyway, you know what I mean? It says the binomial four N plus one is equal to 7 for some value of N please don't solve for the value of N don't do it. Like cheating. Snowfall. We're going to use mindful manipulations. And look for structure to figure out what the value of this complicated expression is. What does this expression equal? If foreign plus one is equal to 7. So what I'd like you to do is play around with this. See if you can use the fact that four N plus one. Is the same thing as 7, right? That's what it means. Without solving for N I don't want you to figure out what N is equal to. Don't do that. All right? Try to use mindful manipulations to figure out what that expression is equal to. Okay? Pause the video and take as much time as you need to. All right, let's play around with it. Well, if we do what we did in the last problem and see that four N plus one is common to both of these, then we'll just pull that right out. What are we left with? Well, we're left with three N plus one. And we're left with N plus two. Thankfully, we're not left with subtraction. So let's clean this up a little bit. Because we're not left with subtraction. We can effectively get rid of both sets of parentheses. And hopefully you're comfortable enough now that you can say, well, three N and N is four N plus one plus two is plus three. Well, okay, I mean, I know what four N plus one is, but what's four N plus three? Well, this is where we want to do a mindful manipulation. And four N plus three will be the same as four N plus one. Plus two. In other words, I can break that three into a one plus two. This is kind of cool because I can now use the associative property of addition. To rewrite it like this. Now why would I do that? Well, I would do that because every time I see a four N plus one, I can put in 7. So what do I have? I have 7. Times 7 plus two. And that 7 times 9, and that's 63. And that's it. That's what this complicated expression is equal to. It's pretty neat, huh? All right, so we're going to clear out all this text, write down what you need to. All right, here we go. All clear. All right. So in today's work, what we primarily concentrated on was treating a binomial as just one quantity, and then factoring it out of a complex expression, all right? We're going to be using this later on when we do other factoring and equation solving techniques, so it's important. To kind of have this idea down. Anyhow, I want to thank you for joining me for another common core algebra one lesson by eMac instruction. My name is Kurt weill. And until next time, keep thinking. That keeps solving problems.