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Christopher Cruz
Christopher Cruz

The Remainder Theorem Common Core Algebra 2 Homework Answers [EXCLUSIVE]

Hello, I'm Kirk Weiler, and this is common core algebra two. By E math instruction. Today, we're going to be doing unit ten, lesson number 11. Known as the remainder theorem. The remainder theorem is one of the most famous theorems in all of rational algebra. And it's an interesting little theorem and kind of takes you by surprise. And we're going to take a look at it and develop it in exercise number one. One thing that would be helpful is making sure that you have your graphing calculator out. Not so much because we're going to be graphing things, but we're going to be evaluating polynomials and that graphing calculator is going to make it very easy to do that. So make sure to have that out. I'll be giving you a little bit of instruction on it. So let's dive right into it. In exercise one, it says consider each of the following scenarios where we have P of X divided by X minus A in each case, simplify the division using a polynomial long division, then evaluate P of A. All right. So in other words, let's take a look this first thing. I want to simplify this division. I'm going to write it in quotient remainder form. And then I'm going to evaluate not P of two P of negative two, but P of two. The way I want you to think about it is if we looked at this denominator, and we figured out where it would make it equal to zero, all right, that's what we're going to be evaluating in here. Okay? Let me just delete that really quick. And maybe switch over to blue. All right, let's do the polynomial long division. All right, let's write this in quotient remainder form. All right, let's divide by X minus two. All right, so from the last lesson I put an X there and match it up, I have an X squared minus two X again, I have to subtract here. I'm going to change it. To addition and add, and that's going to give me a negative 6 X plus 18. Now I have to match up that negative 6 X so I'll put a negative 6 there. That'll give me a negative 6 X plus 12. Again, if it makes it easier for you, change and add, and that will give me a remainder of 6. So all of this will be X -6 plus lots of 6s here, 6 divided by X minus two. All right? So we have a remainder of 6. All right. Now what I'm going to do is I'm going to take that polynomial, which is in the numerator. And I'm going to evaluate it at X equals two. All right, this is where I think that the graphing calculator is very helpful. So just for a moment, let's bring out the TI 84 plus. All right. I'd like to review with you how to use the store command on the TI 84 plus to quickly and effectively evaluate an algebraic expression for a particular value of X so I want to evaluate this quadratic when X is two. So what I'm going to do is I'm going to type in two, then I'm going to hit the store button, which is right down here. I'm going to hit that store button, it's going to put a little arrow there on the screen. And now I'm going to type in X and I hit enter. One of the weird things is then the calculator says two, and all it's really telling me is that, yeah, I stored two in X now what I'm going to do is I'm going to type out that quadratic expression. So I'm going to type out X squared -8 X plus 18. All right, typed it all out. And now I'm going to hit edit. And that 6 that's sitting there. Is exactly. That polynomial evaluated when X was equal to two. So the numerator, when we evaluated it to is equal to 6. Big deal. Whatever, right? Let's do another one. How about this? How about you pause the video now and do the polynomial long division on that one? Okay? Pause the video now. Okay, let's go through it. I've got X squared minus two X -25. And I'm going to divide that. By X -7, I'll have an X here, I'll give me an X squared -7 X again, change negate everything and add makes it easier than I'll have 5 X -25. Plus 5. That'll give me 5 X -35 and change in add and I'll get a remainder of ten. All right? So all of this would be X plus 5. Plus ten. Divided by X -7. So the remainder is ten. All right, what I'd like you to do really quickly is now come over here and evaluate P of 7. Okay? So again, think about the value of X that makes that denominator equal to zero, evaluate P of 7. Do that now. All right, well, whether you store some other mechanism, it's not too bad, and we get ten. Now, what's going on here? Something seems a little strange. You see anything panning out yet that seems a little bit odd. Think about this for a moment. Um. So far, when we evaluate the numerator for that critical value of that denominator, the value of the denominator that makes it equal to zero, we keep getting the remainders. That's kind of cool. All right, now let's see if that keeps going. Let's do a couple that are a bit different. Pause the video now and think about this. All right, now I'm going to clear out the text. Let's take a look at letter C all right. No big deal. There's a little bit of a typo here. It may not be a type on your worksheet, hopefully it's not. This should say P of negative three. Now, let's take a look. Let's do two X squared plus 11 X plus 11. All right, I'm going to divide by X plus three. So I'll have a two X here, and that'll be a two X squared plus 6 X I think I can just do the subtraction here, two X squared minus two X squared, 11 X -6 X is 5 X plus 11. I have to have plus 5. Then that'll be 5 X plus 15, watch myself subtract 11 -15 is negative four. Now, I want you to understand, I'm now going to evaluate P of negative three, because again, I want to look at this as X minus negative three, right? It's always X minus a another way to think about it is that we are going to evaluate the numerator at the zero, the zero of the denominator, the value of X that makes that zero. Pause the video now and evaluate P of negative three. And oh my goodness, look at that. It's negative four. Wow. All right, let's take a look at it one more time. In letter D, why don't you do the whole problem? Okay? Okay, let's go through it. And our pattern continues to work just fine. We're discovering the remainder theorem. But let's go through it. Do the division by X plus four. I have to have a three X there. It gives me a three X squared plus 12 X when I subtract those two. We cancel here I'd get a negative 5 X -20. Therefore, I'd have a negative 5 here, negative 5 X -20 O. Change those. I get a remainder of zero. And when I evaluate P of negative four, I get a remainder of zero. Now this one is a very, very, very special case. Because keep in mind what we just found was we just really found the fact that this thing has to be X plus four times three X -5. And we had a remainder of zero, right? There it is. That may be the most important case of all, right? If we know that that polynomial ends up having a value of zero, at X equals negative four, then X plus four must divide into it nicely. It's got to be a factor of it. All right? Whenever the remainder is zero, then the quantity that's dividing into it has to be a factor of that polynomial. But now I think we can summarize, we can summarize the remainder theorem. First, though, pause the video and think about this slide. Okay. Let's summarize the remainder theorem. It's really quite cool. The remainder theorem. When the polynomial P of X is divided by the linear factor X minus a now, again, notice it's a very simple. Very simple linear factor, X minus a, not two X minus a, not 5 X minus a, not X squared minus two X, then the remainder will always be P of a. In other words, P of X divided by X minus a is equal to some other polynomial. Some other polynomial. Plus the polynomial at a divided by X minus a really, really, really critical the remainder theorem. And let's apply it. All right, this is a very, very simple application. This would be one of those things where it's just do you understand the remainder theorem. It says if the ratio X squared -11 X plus 22 divided by X -9 was placed in the form, Q of X equals R Q of X plus R divided by X -9, where Q of X is a linear function, then which of the following would be the value of R now one way of doing this is just going through the polynomial lawn division. So one way of doing this is just using what we did in the last lesson. You just do this. And do this, and I'll do those X squared -9 X, maybe change it, add negative two X plus 22, then I'd have negative two, and I get negative two X plus 18, and maybe I subtract, and I'd get four. So there would be a few choice four. But what the remainder theorem says is, I don't even need to work that hard. All I need to do is take this, that polynomial, X squared -11 X plus 22, and I just need to evaluate it at 9. All right, I'm going to do this kind of long hand so that we have some faith in how it's working. All right, it would be 81 -99 plus 22. Which would be 103 -99. And that would be four. Right? So it's remarkable that if you evaluate the numerator at the zero of the denominator, that gives you the remainder upon division. Really remarkable. We're not going to delve deeply into why that is the case. We'll leave that for your next course in mathematics, pre calculus, or whatever you want to call it. But it is remarkable that that works. Pause the video now. And think about this problem. Okay, let me clear out the text. All right, by definition, now this is very important problem. By definition, X minus a is a factor of P of X if P of X divided by X minus a is Q of X so in other words, something is a factor of a polynomial if when we divide the polynomial by that factor, we get another polynomial. Hopefully not the same one. All right. This is like saying that three is a divisor. Whoops is whoops. I don't know how that got down there. Three is a divisor. Of 21. Because 21 divided by three is 7, right? It's not a fraction here. It's a whole number. Same idea here. So by definition X minus a is a factor of P of X if P of X divided by X minus a is something nice again. Another polynomial. What must be true of the remainder IEP of a for X minus a to be a factor of P of X explained. So what's the remainder gotta be? If X minus a is a factor of P of X. Well, let's take a look. I mean, we just had the remainder theorem that said, wait a second. P of X divided by X minus a isn't just Q of X, it's Q of X plus the remainder divided by X minus a but we're saying that if X minus a is a factor, this is true. Well, then R must be equal to zero. The remainder must be equal to zero, and this is very important. Thus, P of a must be equal to zero. And this gives us a wonderful way of knowing whether a simple factor, like X minus a, or a simple, a simple expression like X minus a, is a factor of a more complicated polynomial. If the polynomial is equal to zero, at the zero of that factor, well, then that factor, and that expression is a factor of the polynomial. Sorry, not saying that that well. Let's jump into it and see how this can be this can be used. Pause the video now, write down anything you need to. And then we'll take a look at how to use this. And maybe I will. Maybe I won't be so confusing. All right, exercise four. Determine if each of the following are factors of the listed polynomials by evaluating the polynomials. In other words, without trying to factor it all is X minus three a factor of this polynomial. Well, it's this simple. It's going to be a factor if P of three is equal to zero. All right? And let me do this the long way. By hand, P of three, three squared -11 times three plus 24. Let's see, that's going to be 9. -33 plus 24. And that's negative 24 plus 24, and that's equal to zero. So yes. It is. X minus three is a factor. Of that polynomial. How about X plus 5? Is that a factor of this polynomial? Well, for that, what we're going to want to do is not evaluate P of 5, but P of negative 5. Okay? And again, let's kind of do this the long way. And then we'll have you do the last two on your own, and we'll take a look. Right? So that would be two times 25. -45 minus two. Gives us 50 -45 minus two, and that's equal to three. Three doesn't equal to zero. So no. All right. The only way X plus 5 would be a factor of that polynomial is if P of negative 5 was zero. So why don't you try a couple that involve cubics? Use your calculator to evaluate the polynomials that's absolutely fine. Okay. So let's go through them. That's a little bit weird that letter D is over here, letter C is over there. But hey, why not? So is X plus one a factor of this? Well, if it's a factor then P of negative one should be equal to zero. And in fact, if we evaluate P of negative one, we do find it's equal to zero, and therefore yes, X plus one is a factor. In this case is X -5, a factor of this cubic, well if it is, then P of 5 would have to be equal to zero. And if we evaluate that, we find that P of 5 is equal to negative 5. So no. It's not. A factor. All right. And we kind of already had a sense for this. Maybe in the last unit. We looked at graphically factoring something. When we were working with polynomials. And we knew the zeros. And we use those to kind of backtrack the factors. And it's really the same way here. We're testing to see if something is a zero. And by knowing whether or not it's a zero of the polynomial, we can know whether some other thing is a factor. Okay, pause the video now and write down anything you need to. Clear out the text, and then let's do one last problem. All right, exercise number 5, for what value of K will X minus four, the factor of X squared plus KX -52. Show how you arrived at your answer. All right. Pause the video now and think about how you might be able to solve this problem. All right. Well, let's go through it. The plain fact is for X minus four. To be a factor, X equals four must solve the following equation. X squared plus KX -52 equals zero. X equals four must solve this. It must be a zero of that. All right? Well, if it's going to solve it, then I can substitute it in. I can put a four in here. And I can put a four in here. And the beautiful thing is that the only thing left is the is the variable K, or what I'm trying to solve for. So I have 16 plus four K -52 is equal to zero. Which would give me four K -36 is equal to zero. Four K is equal to 36. And K is equal to 9. That's the only value of K that will make X minus four a factor. What's really kind of cool is you could then go back here and you could say, oh, there I have it. I wonder how to actually factors. And the cool thing is it would factor like this. All right, and that negative four times 13, a little bit maybe. Maybe trickier gives me negative 52. But yeah, as long as there's a 9 there, X minus four will be a factor. But the key to doing that problem easily is knowing that X equals four would have to be a zero. Okay, pause the video now, and then we'll wrap up the lesson. All right, let's do it. So today we saw one of the most remarkable theorems in rational algebra. What was known as the remainder theorem. And the remainder theorem is kind of hard to state without writing stuff down. Basically says if we take a polynomial and we divide it by a simple linear factor. Then the remainder that we get will be can be determined by evaluating the polynomial at the zero of that factor. The zero of that, that divisor. This is see, I told you it wasn't going to work well. One of the nice consequences of that is that we can easily know whether or not a particular binomial is a factor of a particular polynomial by evaluating the polynomial at the zero that that that binomial quantity gives us. Not easy to do without writing stuff down. All right. Well, thank you for joining me for another common core algebra two lesson by E math instruction. My name is Kirk Weiler, and until next time, keep thinking. I keep solving problems.

The Remainder Theorem Common Core Algebra 2 Homework Answers

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