 Welcome to our review for exam 3 for math 1050 college algebra at Southern Utah University. As usual, I'll be your instructor today, Dr. Andrew Missildine. This is not our first exam. Clearly, this is exam 3. We probably know how the exams are structured for college algebra by now. I won't delve too much into the structure of the exam. Mostly, I just want to talk about the topics that you're going to see on this exam, because as the structure and organization's exam, the resources available to you as you study, these are basically going to be the same as the previous exams, and the time, place and manner of exams does change from semester to semester. So do check with your instructor or with Canvas to get the exact dates for those things. There'll be two types of questions on this exam, multiple choice, and free response like the previous two exams. You will see nine multiple choice questions. They're worth five points each. And then we will see six free response questions, which range anywhere from five points to 10 points. And so the topics of this exam are basically going to cover polynomial functions and rational functions. So the topics we have seen in chapters four and five in our lecture series, in particular, that'll cover lectures 22 through 35. Now, I should mention that lecture 36 is actually in chapter five for our lecture series. This was the lecture about partial fraction decompositions, mostly to do with timing of the exam and such. This one section will not appear on exam three. It will appear on exam four, so we do need to know it. But just so you're aware, partial fraction decomposition, even though it is part of chapter five, it doesn't appear on this exam. It'll be on the next one. So without further ado, let's get into the topics of this exam. Nine multiple choice questions. Like I said, the first question is going to be a question about multiplicity of roots of polynomial functions. So you'll be given a polynomial function, and it'll probably have a very high degree. Like we're looking at this one, three and two, and one and five. Yikes. That's pretty big. You don't have to graph this one. It will be given to you factored. You have to identify what are the x-intercepts and what are their multiplicities. But it won't be phrased in that way. Like in this question, you just ask where on the graph does it cross the x-axis? All right. So remember that crossing the x-axis is equivalent to an odd multiplicity. And when you touch the x-axis but don't cross it, that's the same thing as an even multiplicity. So even though the question is phrased with regard to the graph, you should understand that multiplicity is giving this information. So you're looking for those ones which have odd multiplicity or a variation of this question might have to do with even multiplicities. So this idea of multiplicities of roots of rational functions, this was introduced in section 4.3 in our lecture series. In particular, I'm just going to give the lecture numbers. That's probably going to be easier to look up. You're going to want to consult lecture 24 for some more examples there. Question number two is going to give you a rational function. It might be factored like this one. It might not be on question one. If this thing's going to be huge, you're going to be given factored form because it would be too difficult otherwise. Not impossible but difficult for a multiple choice question. It's given to you factored. But on question number two, it might be factored. It might not be. It doesn't make much of a difference. You're going to want to find the horizontal asymptotes for, if any, for this rational function. I should say, I said asymptotes, but admittedly, if it has a horizontal asymptote, it will be unique. So really, find the horizontal asymptote, if any. But I throw it out there as a plural possibility because vertical asymptotes can show up multiple times. A horizontal asymptote, if there's any, only shows up just the once, right? So how do we find these things again? Remember, the idea is to look at the leading term on top, the leading term on top, and then you look at the leading term on the bottom. And then you want to simplify and you'll get something like 1 over x squared or maybe like 3x cubed. You're trying to get either a monomial or a reciprocal function when you're done. In other words, you're looking for this idea about, is it top heavy? That is, is the power on top greater than the power on the bottom? In that case, of course, you're going to get no horizontal asymptote. There could be an oblique asymptote or something else, but the in behavior will look like a polynomial. This question is only asking for the horizontal asymptotes. It could be bottom heavy. Remember, that's when the denominator has a larger degree than the numerator. In that case, your horizontal asymptote would be y equals 0, right? It's just the x-axis. Now, I want you to be aware that no matter what rational function you'll be given, one of the options will be y equals 0 because that's the bottom heavy case. And then there will also be an option of no horizontal asymptotes because, again, that's the top heavy case. But you could also have the balanced case where the top and bottom have the exact same degree, in which case you're going to get y over some a and b, where a is the leading coefficient on top, b is the leading coefficient on the bottom. And do make sure, assuming this was balanced, you might have to multiply these out three times two, one times five, something like that. And so that's where the other four possibilities come from. Possibilities if it was a balanced rational function. Horizontal asymptotes were first introduced in lecture 30. In particular, section 5.1 was about the asymptotes of rational functions. But as we were graphing rational functions in lectures 31 and 32, sometimes we had to graph rational, excuse me, as we're graphing rational functions, we had to graph horizontal asymptotes. So you can find some examples in section 5.2 if you want some more there. Question number three is going to be a question about power functions. In particular, you're going to want to solve an equation using some type of rational exponent of some kind. So this is something we introduced in the very first section for this unit, lecture 22 covering section 4.1 about power functions. So how do you solve something like this? There's a few things to remember, of course, that if you have x, x to the a, b is equal to some other number, it's called y, right? Then you're going to want to take the reciprocal power of both sides. So this will translate to bx is equal to y to the b over a. Now, again, like I said, there's a few things you have to watch out for in this situation. So for example, if the original value of the original numerator here, if this was an even power, so like two thirds, two fifths, four sevenths, something like that. If the numerator is even, when you take it's reciprocal, that means the denominator here is going to be even. And that actually suggests that this is going to be a square root, an even root of some kind. So you have a square root in play. As such, when you take the square root to solve an equation, you're going to have to take a plus or minus. So there might be two solutions to the equation because essentially you have to take a square root. So that's something you definitely want to watch out for. Now, of course, if this fraction is not in lowest terms, you want to reduce it so there's no common divisors between numerator and denominator, two thirds is already in that case. And you'll most likely be given that, but watch out for it just in case that is the situation, all right? Another thing you have to watch out for is if the denominator of this exponent was even, right? Because in that situation, that means you have a square root already in play or maybe a fourth root, same basic thing in that situation. And so you have to ask yourself, is this y value non-negative? Because I want you to watch out for something like the following. If I take x to the three halves and this is equal to negative one, this is the same thing as saying the square root of x cubed is equal to a negative one. And square roots, much like absolute value, cannot be negative. That's outside the range of the function right here. And with therefore, there's no real number whose square root is a negative. Therefore, then you get no solution in that situation. So those are some things to watch out for. This one is usually fairly easy. This one's a very common mistake for students. They forget the plus minus. So do remember to do that when you're solving these power equations. Now, I started scribbling on the next problem, so I do need to erase it a little bit so I can go on down. So looking at question number four, on this question, you'll be given a polynomial function. Most likely it'll be expanded outward. You won't be given a factored. In which case, you're going to be asked for the possible rational roots of that polynomial. So this is what we did, what we learned about this, among other things, exactly in lecture 26, coming from section 4.5. You might also want to consult lectures 27 and 28. But in lecture 26, we learned about the rational roots test for which, remember, we look at the constant term of the polynomial. We look at the leading coefficient. So we get these four over three. And we have to consider all the possible divisors of four thirds, both positive and negative. So we look for all the possible divisors of four, all the possible divisors of three, and that will then give us a list of all the possible rational roots. And so the rational roots test is what you want to use in this. So the rational roots test was given in lecture 26, but in lectures 27 and 28, we used the rational roots test, among other methods, to factor large polynomial functions and use that to solve polynomial equations. So you can get some extra practice of the rational roots test by looking at 27 and 28, but 26 is where it originated from in our lecture series. So then coming to question number five, this is similar to question number two, but now you'll be looking for an oblique asymptote. An oblique asymptote occurs exactly when the numerator is exactly one degree larger than the denominator. So in this situation, you have a degree four over degree three. So it is top heavy, but because they only differ by one, asymptotically, this rational function is the same thing as a line, mx plus b. And how do we find that line? Well, we're going to do this by long division of polynomials. So you would take 3x to the fourth minus x squared and divided by x cubed minus x plus one. We need to find the quotient up here. We could care less about the remainder in this example because the asymptotic behavior, that is the oblique asymptote, is the quotient. The remainder goes off to zero as x approaches infinity in this situation. So just as a reminder, section 5.1 is when we introduced asymptotes, including oblique asymptotes. So again, look at lecture 30 for some examples of this. Of course, you can turn also to lectures 31 and 32 as we are graphing rational functions in those lectures. There was an example or two where an oblique asymptote showed up, so you can find some more examples there. So the last question on page number one of the exam question 6, this is a question coming from section 4.5 about considering complex roots of a polynomial. And so in which case you're going to want to use the fundamental theorem of algebra and its corollaries here, that if we have a degree 3 polynomial and it has real coefficients, if we know that 3 is a root and we know that 1 plus i is a root, can we come up with that polynomial? In which case remember, if you have a complex root with a real coefficient, then its complex conjugate has to also be one of the roots. And that, then you should count and it gives you all the roots 3, 1 plus i, and then the other one, right? The other thing to remember by the factor theorem is that if c is a root of your polynomial f of x, then we have that x minus c is a factor of f of x. So this information about factors, excuse me, this information about roots turns into information about factors, and then we can put together the polynomial, multiply it out, and you don't be too crazy, maybe you have 3, 4 roots, something like that. In which case then we select which of these real polynomials have these as the correct roots, all right? Moving on to the next page, let's finish up the multiple choice section. Question number 7 is going to have to do with the binomial theorem. So you have 2x plus 1 to the fourth. Now you could try to multiply it out the hard way, but the binomial theorem is going to be much faster and less likely to have errors in it. You just have to compute the coefficients, of course. These just are the binomial coefficients, but you can also use Pascal's triangle here, 1, 1, 1, 2, 1, 1, 3, 3, 1. Since we have to take the fourth power, we want the next row, 1, 4, 6, 4, 1. Remember, you just add together the numbers, 3 plus 3 is 6, 1 plus 3 is 4, go from there. And so these numbers give you the binomial coefficients. If these numbers have some coefficients, like there's a 2x there, you have to pay attention to that if there's a plus or minus sign of pay attention as well. But that will then help you, of course, multiply this thing out. We learned about the binomial theorem in section 4.6, in particular that was lecture 29 in our lecture series here. Moving on to question number 8. Question number 8 is going to ask about the domain of a rational function, but it's going to ask in a couple of ways, right? It might ask you, here's a rational function probably factored. What's its domain? The domain is everything that makes the polynomial undefined. That is the domain, I think I said that backwards, I'm sorry. The domain will be everything except which makes the rational function undefined. And as such, for a rational function, the one problem we have to look out for is division by 0, right? What makes the denominator go to 0? When the factored form, you should be able to recognize these things very quickly. So the domain will be everything except for what makes the denominator go to 0. That's the domain. Now, this question might ask you something different. It might ask instead about the vertical asymptotes of the rational function. When you are looking for the domain, right? When you're looking for the domain of a rational function, you use the unsimplified form. You don't cancel any common divisors if there are any on top and bottom. This example doesn't have any. But if there was a common factor, like you have x times x minus 1 over x times x plus 2, right? You attempt to just simplify this thing out. If you want to find the domain, it's like, uh-uh, the domain would be everything except for 0 and negative 2. It doesn't matter if things cancel out, because since there's a x on the bottom, you're going to divide by 0 if you plug in x equals 0. You actually end up with 0 over 0, but that still is not a real number. Watch out for such a thing. Now, when you're looking for a vertical asymptote, in that situation, you do want to look at the simplified form. It could be, oh, I think I spelled simplified wrong, right? Unsimplified. To find a vertical asymptote, you need to look at the simplified form so you do cancel these things out. For this one, while the domain is everything except for 0 and negative 2, there's a vertical asymptote at x equals negative 2. What happened at x equals 0? That actually gives us the third possibility for this question. This question might ask about a remove point. Does this rational function have any remove points? And if so, where are they? The remove points happen at the canceled part, right? So the domain, you look at it unsimplified. The vertical asymptote, you look at it simplified. And for the remove points, you look at the cancel part. So those things that cancel out actually were remove points. So this thing has a remove point at x equals 0, like so. And so question number 8 is going to ask you something in this range, which these are all related to each other, the domain. The vertical asymptote is something, is an exception to the domain. A remove point is an exception to the domain, right? In which case you might have to differentiate between the two or you might have to put them all together. So for question number 8, again, go to lecture 30, section 5.1, about asymptotes of a rational function, including vertical asymptotes here. And that's what you're going to want to study for. So make sure you're very careful on this one. Question number 9, the last little choice question will ask you to solve a polynomial or rational inequality. These were all covered in lecture 33, polynomials and rational inequalities. Be prepared to solve them. You'll have to solve it using interval notation, of course. So something like that. All right. So then moving on to the free response section, we see question number 10 here. Can you compute the quotient and remainder of a rational function given right here? This is, you're going to want, well, basically, it's demanding that you do a polynomial division, long division of polynomials. Synthetic division is not going to work here. I mean, if it was available, you could use it, but I'm not going to give you a linear divisor. Something quadratic or bigger is going to be given to you. So you need to compute the quotient and remainder. Do describe who they are specifically using long division of polynomials like we did in lecture 25. Section 4.4 was about polynomial division. This is worth seven points. The next question, number 11, you'll be asked to solve a radical equation. Find all real solutions to this. Radical functions we talked about in lecture 35. That is section 5.5. Amongst the things we learned about rational or radical equations, radical functions, we learned how to solve radical equations. And one thing you have to be very cautious about is this idea of party crashers, right? These fake solutions that show up in a equation. This is a very big problem for radical equations. You can solve this equation and you might end up with two solutions like x equals a and x equals b. You need to check to see if they actually solve the original equation. If one of them doesn't, it needs to be removed. Maybe none of them do, in which case there's no solution. Maybe they both work. It kind of depends on the problem itself. And honestly, there's really not a good prediction I can give to you in this review. So the idea is you just need to check. Even if both answers work, if you do not check your solutions, you can't get full credit on this problem because it's necessary to check radical equations because of these party crashers. And the main idea, the main reason for that is that while the square root is a one-to-one function, x squared is not the pseudo inverse of the square root. And so one has to be careful that when you take the square root, you're potentially entering into the problem new solutions that actually don't work, right? Like if x equals two and you square both sides, you end up with x equals four, which actually gives you two solutions, two and negative two. We have to make sure that we account for these party crashers. And it's not necessarily the negative number. Negative numbers actually can work. You have to check both of them. So watch out for that. Question number 11 was worth six points. Question number 12 was worth seven. This one also comes from lecture 35 about radical equations. Because remember, radicals are the inverse functions to monomials. So like the square root is the inverse of x squared, if we ignore one-to-one this and things like that. The cube root is the inverse to x cubed and such. And then of course, this radical one, we also consider fractions and things like this. So if you're giving a rational function like this, negative two x cubed over x cubed minus one, how do you find the inverse function? Well, this is something we studied before, but now we're turning up the heat a little bit, right? So you have y equals negative two x cubed over x cubed minus one. You need to switch the x and y. So we get x equals negative two y cubed over y cubed minus one. And then you solve for it. Now in this situation, you're guaranteed to have something like a y cubed or a cube root of y or some power of y. You're going to want to solve like in this one, you want to solve for y cubed. And so you'll get something like y cubed equals g of x or something. And then at that moment, you can then take the cube root, y equals the cube root of g of x, which is probably going to be a rational function at that point. And so there's going to be some exponents in play here, but that's okay. The exponents not going to really complicate things too much. This question number 12 is not fundamentally different than the corresponding question we saw on exam one. We have to solve for inverse functions. And so if our polynomial function or rational function is invertible, that is, if it's one to one, we can solve for that inverse function and it'll typically involve a radical of some kind. Or if we started off with radicals, then we'll probably get a monomial when we're done. Question number 13, this question is sort of like half free response, half multiple choice. You will be given a polynomial function probably of high degree. So this one's like degree six. You're asked to use synthetic division. You're going to divide by x minus c, where c was provided up here earlier, it's negative three. In which case, then you want to first do synthetic division. So this is how you have to show your work. So you're going to show your synthetic division here. So like two, zero, negative 18, zero, one, negative nine, excuse me, zero, negative nine. So of course, you need to make make sure there's placeholders. If there's any zeros in there, like there's no x to the fifth term. Then that means it needs to be a zero in your synthetic division. We're divided by negative three. So go through all the business. So you go through and fill all that stuff in. So that is necessary for full credit. And then you have to analyze the synthetic division. So you get a remainder at the end. What does that tell you about the root? If you have a remainder of zero, that means it's a root of the polynomial. If you get something other than zero, that means it's not a root of the polynomial. So you know, say something like that, then can we say something about being an upper bound or lower bound? If your root is positive and the entire denominator, excuse me, the entire bottom of the tableau is positive, or when I say positive, it's just never negative zeros. If the bottom row is never negative, if it's non-negative, we'll just say it's always positive for simplicity. And you start off with a positive number. That means it's an upper bound. As this was a negative number, negative numbers we can never detect as upper bounds, but they could be a lower bound. If you have a negative number and you get an all-trained sign plus, minus, plus, minus, plus, minus, which again, zero is like a wild card. It could be plus or minus, depending on whatever is best. You could get a lower bound. So consult with the upper bound and lower bound theorems for question number 13. Remember, a positive root can be an upper bound, but it can never be a lower bound. It could be neither, of course. A negative root can never be an upper bound, but it could be a lower bound or it could be neither. So you definitely need to circle your answers there in order to get full credit on question number 13. All right. Now we move to the graphing page for this review. Questions number 14 and 15 are going to be about graphing. Question number 14 will be about graphing polynomial functions like we did in section 4.3, namely lecture 24. What do we look for when we graph polynomials? Now, by virtue of my kindness, this polynomial will be factored, so we don't have to go through the effort of factoring it. But there's some things we need to look out for. We need to look for the x-intercepts. The x-intercepts are going to come directly from the factors of this thing. We have to also pay attention to their multiplicities. So we'd say something like, oh, x equals whatever is a root with a multiplicity of two. That's an even multiplicity. So we'll touch the x-axis. Then we could say like, oh, x equals yada, yada, a different number has a multiplicity of three. That's an odd multiplicity. So it's going to cross the x-axis. We're going to scribble that stuff right here. So show your work by listing your x-intercepts, your y-intercepts. I also want you to see the leading term, right? The constant term is the y-intercept. So looking at all the coefficients, the constant coefficients, you get the y-intercept. Pay attention to powers. To find the leading term, of course, you also look at the coefficients. But now you look at the powers of x, including the powers. And you're going to write that here. The in behavior, let's say hypothetically your in behavior turned out to be something like negative 3x to the fourth. Remember, that means you're like, oh, you point down on the right. You point down on the left. That in behavior needs to be listed. And you can show your work by computing the leading term. Okay. So then you would label the x-intercepts that you have discovered. Again, this is all hypothetical. And then connect the dots, something like this. Then your in behavior should match up with what you had over here. The multiplicities, you have a touch, you have a cross, you have a cross that should mention over here. This thing does not need to be drawn to scale. You'll notice that there's little tick marks on the x-axis, but there's none on the y-axis. It could be that your y-intercept turns out to be like 5, or 50,000 or something like that. This does not need to be drawn to scale. I just want to see the basic shape of the polynomial. A bad picture is actually the correct answer for a question like number 14. It doesn't have to be drawn to scale. Question 15 will be very similar, but now we're going to be looking at a rational function. Graphing rational function, we actually spent two lectures on this one, because there's a little bit more going on here. Lectures 31 and 32 covering section 5.2 from our lecture series. So some of the same things are going to be present here. You have to look for x-intercepts and y-intercepts. Of course, there's only going to be one y-intercept if x equals zero is inside the domain. For the x-intercepts, you do need to mention the multiplicities. Same thing. So if it's like, oh, this number has an even multiplicity of two, that touches the x-axis. This one has a multiplicity of three across the x-axis. I need all that information listed over here. Same thing with the y-intercept, but you don't mention there's no multiplicity for the y-intercepts. You'll also have to mention any acetotes. Is there a horizontal acetote? Is there an oblique acetote? You need to include those things. If the graph approaches y equals zero, that should be indicated and labeled. If it approaches infinity, that should also be labeled and indicated. Right? Maybe there is no horizontal acetote. You have to say that. Are there vertical asymptotes? The vertical asymptotes, we're coming from what makes the denominator go to zero. You have to also watch out for remove points. If there's a common factor on top and bottom that cancels out, that's a remove point. But otherwise, you get vertical asymptotes. You need to list those, like x should not be this number, x is not this number, whatever. But mention there are asymptotes, are there the multiplicities of the asymptotes? So if you have an even multiplicity, it's like, oh, this acetote touches infinity, but doesn't cross. Oh, this one has an odd multiplicity. It crosses infinity and things like that. So then you'd graph these things. You have your point. Here's an x-intercept. Here's an x-intercept. Let's say there's a horizontal asymptote. So let's say there's a y-intercept. Let's say there's some vertical asymptote right here and right here. So we might then put it all together and say something like the following. We go off towards infinity, then we touch infinity, we cross the x-axis, go off towards our horizontal asymptote. So then we come up, we touch the x-axis, come back down, and then we wrap around infinity and do something like that. So you know that could be the picture we get, right? In which case it's like, oh, we touched the x-axis because we had an even multiplicity. We crossed the x-axis because we have an odd multiplicity. We touch infinity because we have an even multiplicity for the asymptote. We crossed infinity because we had an odd multiplicity for the asymptote. The y-intercept is given here. The horizontal asymptote is given here. If this is even the information that's on the graph, this is just a hypothetical situation. Be prepared to graph these things. Both this question, this question's worth 10. The previous one is also worth, excuse me, 8 points. Polinomials are a little bit easier because you don't have to worry about the asymptotes whatsoever. And so now we arrive to the last question of the exam, question number 16. This one's worth also 10 points. So that is the maximum of any question on this test. This one you'll be asked to find all of the real or complex zeros, a.k.a. roots of the polynomial, right? So what are the things that make this thing go to zero? What are the roots of this polynomial here? And so this one definitely is a wildcard type question. The degree of the polynomial will be large at least the degree three, because if it's just quadratic, you can immediately start with the quadratic formula. And this one we want to use the more advanced factoring techniques. It could be degree four, like we see here. It could be degree five or six. Like I said, it's a little bit of a wildcard there. What can you use? Well, basically all of the tools we have in the toolbox you can use. Some things you definitely want to use will be something like the rational roots test. We had a multiple choice question about that earlier. You definitely want to use the rational roots theorem. So looking at the constant term and lean coefficient, that'll then give you a list of possibilities. You want to use that. You can use Descartes rule, a variation of sine. Descartes rule. So I mean, you can look at the variation here. So it's like it's switched once, it's switched twice. So that tells you if you have either two or no real roots, you can talk about the negative roots as a possibility as well. Certainly it's not mandated. None of these techniques are mandated, but it can be very helpful, especially when you recognize by Descartes rule, there's maybe no positive roots or no negative roots. That can be very helpful in speeding up the process. So once you have your rational roots like P over Q, and maybe use Descartes rule to help you determine how many positive or negatives are, what's the better one? You have all these possibilities. You need to start a little bit kind of guess. I would generally try to do smaller guesses more than bigger guesses, right? Because the biggest possibility here is 10. It's probably not 10. And if 10 was a root, that kind of forces that one or negative one is also a root. So again, you're better off guessing smaller choices. So you're probably, if you're looking for a positive root first, your possibilities could be like one, two, five, or 10. You're better off doing one or two. Maybe try two and then go from there. If two works great, go from there, reduce the problem. If two didn't work, is it an upper bound? Can we use the upper or lower bound theorem? So let's also mention that here. We have our upper and lower bound theorems. So these help us play the hot and cold game, right? Because if it is an upper bound, that means we need to try smaller. If it's not an upper bound, maybe we try bigger. Similar things with negative roots in the lower bound theorem. So we can play that hot and cold game until we find all of these rational roots, okay? Of course, other factoring techniques that we know can apply. So if you find a common divisor, a GCD, you can factor that out. If you can factor by groups, do that. That can be helpful sometimes. A difference of squares, a difference of cubes, a sum of cubes. Those are relevant factorization formulas. The perfect square trinomial is also helpful. Honestly, factoring of quadratics in general, that reverse foil technique we've learned about in Chapter 2, is relevant. You can use the quadratic formula also. But of course, that only helps you get to a quadratic. As this is Degree 4, you're going to have to find two roots before you get there. But again, you're going to use lots of synthetic division on this one because you're going to test roots. That's going to be your most effective way of doing it. You want to test a root, in which case you try something like, okay, let's see, 1, negative 5, negative 5, 23, and 10. Try 1. See what happens there. You get 1. 1 times 1 is 1. Minus 5 is negative 4. Times 1 is negative 5. Times negative 5 is going to be negative 9. Times 1 is negative 9. Add that to be, excuse me, 23 minus 9, of course, is going to give us 14 times 1 is 1. Excuse me, 1 times 14 is 14 plus 10 is 24. So we can see that, okay, that didn't give us a root. But we don't have an upper bound per se. So my next guess might be like 2. And we kind of just play around with this until we find the roots. Now, if you do find a root, remember, this gives you the depressed polynomial. Let's say 1 did work. In which case we get like, oh, one of the factors was x minus 1. And then the other factor was x cubed minus 4x squared minus 9x plus 14. And then you start to factor this thing. Now, again, this didn't work. This is not the correct factorization. Don't worry about that. But you then can use the depressed polynomial to move forward, right? You can make some new analyses, like use the rational roots test again, use Descartes' rule of signs, and then continue to go. Of course, if one root didn't work before, like if one didn't work before and you then factor later, one's still not going to work until we eventually get all of the roots. There could be non-real roots. There could be irrational roots. But you would find those probably using the quadratic formula. I'm not going to give you too many irrational or too many non-real roots because that would be very difficult. Most of the roots will be rational roots, except for maybe 2. In which case those last two you would find using the quadratic formula. And so then all of this stuff, this idea of factoring, this advanced factoring of polynomials, this was the main coverage of lectures 26, 27, and 28. 28 specifically focused on non-real complex roots. The real roots were covered very much in 26, 27. But you want to know all of those things, all of these things from section 4.5 because that's what question number 16 is going to cover. And so that's then the last question on exam number three. And thus that completes our review. If you do have any questions about the exam that were not covered in this review video, please feel free to consult me. You might have noticed one suspiciously missing question from the lecture series that you might have not seen the question for. In particular, you might have noticed that the lecture about polynomial story problems and rational story problems seems to be missing from this video. How suspicious that is, right? In particular, that was lecture 34. There was no question that really covers those story problems. Well, my treat for you for making it to the end of this review is actually a promise that none of the story problems are going to show up on this test. There's plenty of content as it is. And frankly, honestly, if you can do well on the topics covered in this exam, I feel you probably can do okay with enough queuing up for story problems that such a question won't be necessary. All right, so if any other questions do come up as you're studying for this exam, please let me know and I'll be glad to help you. See you next time, everyone. Bye.