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Contents

- 1 Introduction
- 2 The Bay Area
- 3 The Bay Area
- 4 Path of a Function
- 5 Path of a Function
- 6 Consider the y values
- 7 Consider the y values
- 8 Intermediate Value Theorem
- 9 Intermediate Value Theorem
- 10 Train Schedule
- 11 Train Schedule
- 12 Narrowing the Window
- 13 Between Two Integers
- 14 Between Two Integers
- 15 Counting by 0.1s
- 16 Counting by 0.1s
- 17 Counting by 0.01s
- 18 Counting by 0.01s
- 19 Sketch the Graph
- 20 Sketch the Graph
- 21 Function Chart
- 22 Function Chart
- 23 Left and Right Intercepts
- 24 Left and Right Intercepts
- 25 Pick a Value
- 26 Pick a Value
- 27 Which Interval
- 28 Which Interval
- 29 Check which value
- 30 Check which value
- 31 Profit Function
- 32 Profit Function
- 33 Rational Roots
- 34 Rational Roots
- 35 Function Table
- 36 Function Table
- 37 Factor from Zero
- 38 Factor from Zero
- 39 Divide for Athena
- 40 Divide for Athena
- 41 Use the Quadratic Formula
- 42 Use the Quadratic Formula
- 43 Breakeven Points
- 44 Breakeven Points
- 45 All Kinds of Zeros
- 46 Factors Zeros and Degree
- 47 Factors Zeros and Degree
- 48 Degree 4 Factors
- 49 Degree 4 Factors
- 50 Repeated Zeros
- 51 Repeated Zeros
- 52 Non Real Roots
- 53 Non Real Roots
- 54 Find The Quadratic Factor
- 55 Find The Quadratic Factor
- 56 Favorite Red Pen
- 57 Favorite Red Pen
- 58 Other Complex Roots
- 59 Other Complex Roots
- 60 Functions From Zeros
- 61 Functions From Zeros
- 62 Number of Factors
- 63 Number of Factors
- 64 Name the Factors
- 65 Name the Factors
- 66 Simplest Polynomial
- 67 Simplest Polynomial
- 68 The Only Function
- 69 The Only Function
- 70 Amazing Polynomial Function
- 71 Amazing Polynomial Function
- 72 Pull Out x
- 73 Pull Out x
- 74 What Next
- 75 What Next
- 76 Positive Factors
- 77 Positive Factors
- 78 Ideas for Roots
- 79 Ideas for Roots
- 80 Find the First Zero
- 81 Find the First Zero
- 82 Integer Constants
- 83 Integer Constants
- 84 Long Division
- 85 Long Division
- 86 Find Two Zeros
- 87 Find Two Zeros
- 88 Find Two Linear Factors
- 89 Find Two Linear Factors
- 90 Finish off the Zeros
- 91 Finish off the Zeros
- 92 We can find the factors!

In the last few lessons, we've looked at graphs of functions and talked about how they behave. We've seen some zeroes of functions that are integers, and some that are fractions. But, what if we can't find x intercepts using methods that we've already learned? What if, for example, our roots are irrational? How can we estimate where they are? That's what were going to look at next.

It is being a while since we were with Athena. So let us get back to talking about her life. Now Athena lives in California, in the Bay area to be exact. To tell a more about you, here is the Pacific Ocean and here is West of California. Here is San Francisco Bay. It is a pretty beautiful area to live in. And she goes to school in Saint Jose Bay right down here. And all of these little cities lie right along the bay now, sometimes Athena likes to take a non-stop train from San Jose, all the way up to San Francisco, which is this northern city right here. Along the train line, which I have drawn in red right here, are a bunch of other stations. Including Sunnyvale, Mount View, Palo Alto, Redwood City, San Mateo and Millbrae. So, let me ask you a question. If Athena wants to take that nonstop train starting in San Jose and going all the way up to San Francisco, which stations must the train pass through along the way? Remember the red path here is the train track. It doesn't branch off anywhere. it just follows this path exactly.

There are ways you may have over thought this question, but I wasn't trying to be tricky here. The truth is that to go from San Jose all the way to San Francisco, we have to pass through every single one of these stations listed here, since their all along the path. That makes perfect sense. They're in between the starting point and the ending point. So to go from here to here, we have to pass through what's in between.

So, how can we tie this to our graph? Well, the train track is in one piece. The train can't just jump from station to station, because it can't leave this track. Similarly, all of the graphs that we've looked at so far are also in one piece, with no breaks. When a graph is like this one, and has no brakes in it. But, it's, rather, one piece connecting two points together. What can we say about the range of that function? So, looking at this graph over here in particular. What numbers can we say are in the range of this function? Does it only contain the numbers, negative 8 and 4? Does it contain all of the integers between negative 8 and 4? Does is span the entire interval from negative 8 to include the points negative 8 and 4? Or does it contain all the real numbers? Please pick what you think is the best answer.

The answer is, this interval. Negative 8 to 4 with the end points included. If we look along the y axis here, we can see that any horizontal line we drew going from y equals negative 8 all the way up to y equals 4, would pass through this line. In other words, all of the y values that lie between negative 8 and this value of y, 4, down to this value of y, negative 8, and of course our curve has no breaks in it. Then we must have passed through every single y value along the way.

Now, is this also true for this graph? Do the curves you see here pass through every y value between negative 8 and 8?

No. We can see that there's a break in the function right here. And we have this open space right around here, which looks like it's going from about negative 1 just up passed 0. For which no points on the function have those y values.

We've seen now that if the graph of a function doesn't have any breaks in it, then the curve is going to pass through all the y values that lie between the y value of the starting point and the y value of the ending point of whatever region of the graph that we're looking at. This concept has a special name. It's called the intermediate value theorem. Now, let's say that all we know about the two end points of this curve, which I've now labeled A and B, is that the y-coordinate of point A is a negative. And the y-coordinate of point B is positive. If that's the case, then what is the one value of y that we know the curve has to pass through between point A and point B? There's only one that we can be completely sure of. Think carefully about this.

Some point must have a y value of 0 along this line. 0 is the number that separates the negative numbers from the positive numbers. And since we know that y is negative at A, but positive at B, it has to be 0 somewhere in between.

The Intermediate Value Theorem gives us some really fundamental information then, about how to find zeros in curves without breaks in them. We just need to look for regions of the graph where y values switch from being positive to being negative, or from being negative to being positive. In either case, the curve must have at least one x intercept somewhere in that region that we're looking at. This is pretty general information, though. And the intermediate value theorem can actually help us even more. In fact, Athena used it on the train in a very clever way. One day, Athena needed to catch the train in Mountain View. She had been visiting the Udacity office for the afternoon. Now, to find out what time her train would leave, she called her brother, Nicos. All he could tell her was that the train would come to Mountain View sometime between 4pm and 6pm. This information wasn't quite accurate enough for Athena. She wanted to be able to plan her afternoon a bit more precisely. So she looked up the schedule for the train. Much to Athena's disappointment, somehow, the one departure time that was missing from the train schedule was the time that the train would leave Mountain View. We have the times for every other stop listed, and the stops are written in order here, just like we saw on the map earlier. Staring in San Jose and then moving north along the track. Based on this schedule, Athena knows a window of time within which the train will leave from Mountain View. What is that window?

The train will leave the Mountain View station sometime between 4:58 and 5:16, since the Mountain View station is between the Sunnyvale and Palo Alto stations.

In the last quiz, Athena at first, had a two hour window of time in which she knew the train would leave the mountain view station. Niko also told her it was sometime between 4 p.m. and 6 p.m. That was a pretty big window. Then, with more information from the train schedule, she narrowed down that window to this tiny little yellow one, right here. Between just before 5 and just around 5:15. The schedule then, is going to help her guess a lot better when the train's going to leave. And we can imagine that if she had even more information about the trains travel from Sunnyvale to Palo Alto, she could narrow down this window even more. We can use this same method of starting zoomed out. And then, zooming in, closer and closer, to help estimate the location of zeros that don't occur at fractional or rational values of x. These are irrational zeros. So, here on the graph, for example, you can see that y is negative down here. But it's positive up here. We know that somewhere in this general region, there's a point where y equals 0. We can find that exact point by zooming in closer and closer.

Let's pick a polynomial function to play with so we can see how the intermediate value theorem actually works. I've chosen y equals x cubed minus axis and on our x axis. Now which integer values of x does the zero here lie between?

Between one and two. You can see that it crosses right here and these are the integers surrounding that point.

This time, the lines on the x axis of the graph are separating every tenth. In other words, we're stepping up by 0.1 with every line. Considering this, and our zoomed in version here, what values of x does the 0 seem to lie between now?

It lies between 1.4 and 1.5.

Once again, I've zoomed in even further. And now the lines along the x axis, these vertical lines, mark every 1/100th, or 0.01. So, I'll ask you again, with this new version of the graph, and this new level of precision. What values of x does this zero lie between?

The 0 lies between 1.44 and 1.45. Now remember, we started out just knowing that the 0 was between 1 and 2. That's a pretty huge range, considering how precise, or how accurate, we've gotten now. So the more information we have, the better the intermediate value theorem can help us, come up with a really close answer, or a really close approximation for x.

Let's look at another polynomial function. This time, we have f of x equals x to the 4th minus x cubed minus 3. Now, on your own piece of paper, I'd like you to sketch the graph of this function. For the part of the domain where x lies between negative 3 and 3. And please include those end points. Now, did you actually sketch the graph? Please tell me yes or no.

I hope that you checked yes. Sketching graphs on your own is great practice for understanding functions and helping your hand get in the practice of drawing smooth curves. Your graph should have looked something like this.

Now that we have an idea of what this function looks like, let's fill out an x f of x chart for it. Please put in the corresponding values of the function here, for these values of the independent variable.

And here are the values of the function.

Based on the table we made in the last quiz, we can determine intervals in which the two x-intercepts we can see in this part of the graph lie. So, we have one x-intercept on the left, right around here, and one x-intercept to the right, right around here. Please use the values of x from the table to tell me the intervals in which you know that these x-intercepts have to lay. Don't forget the intermediate value theorem. Think about how we have to use that in this problem.

The first sign change in f of x that we see on the table, happens between the x values of negative 2 and negative 1. Since f of x between these 2x values is moving from 21 to negative 1. So, we see that over here on the graph from it dropping from this positive value in f of x to the negative value. That means there must be an x intercept between negative 2 and negative 1. The x intercept up to the right, we can see on the table, must happen between 1 and 2 on the x axis. Since f of x here is switching from negative 3 and when x equals 1 to 5 when x equals 2.

I know that we have seen that there are two zeroes in this part of the graph, but let's just pick one of them to focus us on. How about the right most one? Which we knows in the interval 1 to 2. Now that we have chosen us here to focus us on, we want to start narrowing down the interval that we knows the zero lies between. In other words, we are going to start zooming in on this portion of the graph. To do this, I'd like you to first pick a value of x that lies between that interval, between 1 and 2. Type that here. And then, plug that value of x into the function and write what f of x at your value is equal to. Now, remembering that the function is negative when x equals 1. But it's positive when x equals 2. Narrow down the interval in x in which the zero must lie. So, the value of your function is negative, then the new interval will be your x value and 2. But if the function is positive at your x value, then the new interval will be between one and your value. Take a little time and think carefully about this.

So, I'm going to be diplomatic and just pick the value that lies halfway between one and two. Let's let x equal 1.5. If we plug this into our function, we find that x equals negative 3.75. Now since this is negative and we know that f of 2 is positive, that means the 0 must lie between 1.5 and 2.

In the last quiz, we narrowed down the interval in which zero lies to be between 1.5 and 2. To narrow this down even further, first, test the value of the function when x equals 1.7. Then, determine whether the zero is going to lie in the interval from 1.5 to 1.7 or from 1.7 to 2. Remember that the function is negative when x equals 1.5, and it's positive when x equals 2.

The value of the function is positive when x equals 1.7. So that means we know that the zero must lie between 1.5 and 1.7.

Now, since I'm super picky, as you know. I really would like our interval width to be just 1/10 instead of 2/10 like it is right now. So what value of x should we check next in our function? Please fill that in here. Once you've figured that out, please fill that value in to these two boxes right here, to show the two different possible intervals we would have using that x value. Then pick which region of the graph the zero actually lies in.

If we want an interval width of 1 10th, we need to choose the value that's 1 intervals then are going from 1.5 to 1.6, and then from 1.6 to 1.7. Now, the value of the function at 1.6 is negative. So, that means our zero must lie between 1.6 and 1.7 since we know that the function is positive at 1.7. So, what have we done? Well, we started out with a big graph of this function, this quartic function. And we were able, through testing different values of x and f of x to zoom in on one of the zeros of that function. We've narrowed down it's location to a really small interval. An interval of just 1 10th. That's really tiny, considering we started out looking at the domain going from negative 3 to

Grant has convinced Athena to help him out with some of the the Gleaming Glasses business stuff. He's given her yet another new equation, that he says is going to describe the profit the company will make, if he invests x thousands of dollars in advertising. So we have P of x equals x cubed minus 20 x squared plus 118 x minus 164. Grant wants to figure out how much money he needs to spend on advertising to make some sort of profit, but he also wants to avoid spending too much money, so much that it would drain all of his profit away. In other words, Athena needs to figure out, for what values of x does this equation P of x equals 0. In other words, Athena is looking for routes of this function. We're not quite ready to help her answer this question yet though. Let's backtrack for a second. First off, can you tell me what the minimum number of real zeros this function has to have is? Think critically about what kind of polynomial this is.

And the answer is 1. Since the 2n's of a polynomial function with an odd degree, like this cubic one right here, need to point in opposite directions. Either one pointing to negative infinity, and the other to positive infinity, or vice versa. The curve has to pass through the x-axis at least once.

Let's try to find the zeros then. We can use the rational zero theorem to come up with some ideas for what our zeros might be. So what are all of the possible rational zeros this test tells us about. Please write them in this box and separate them by commas. Don't forget the negative numbers.

To use the rational zero theorem, we first need to find all the factors of negative 164 and positive 1. Well, 1 is easy. We just have 1 and negative 1. There are a few more for negative 164. We have 1, 2, 4, 41, 82, 164 and the negative versions of all of those numbers. Now that we have two sets of factors, all we need to do is take one factor from this batch and divide it by one of the factors from over here. Now, since all the factors on this right, that we are going to be dividing by, are just 1 and negative 1, the only ratios you would end up with are all of the numbers over here. So, there we go, we have the list we were looking for. Even though this is still a lot of numbers, it's way fewer than the infinite number we were thinking about before, when we had no idea what the roots were.

Here's a list then of some of the possible rational zeros that we came up with in the last quiz. In fact this is just the positive versions. Starting at the top of this list and then working down, please plug in each of these values of x into the function. And oops, I made a mistake. I accidentally wrote f of x instead of P of x. Remember it's super important to keep the names of our functions straight so that we know what functions we're talking about, but everyone makes mistakes including me. Once you've found the value of the function at each of these X values please evaluate whether or not it's a zero. If P of X does equal zero then please type O Y in the zero column, but if it doesn't type and N. Since some of these numbers are pretty big I'm going to make your life a little bit easier. As soon as you type a y into the zero column, you can stop. We're just trying to find which x value is the first zero. Or at least the first zero on this list. That will be enough information for us to continue simplifying this function, and figuring out more information that we need.

When x equals 1, I find that P of x equals negative 65. Since this isn't equal to 0, 1 must not be a 0 of this function. Let's check out 2. Sure enough, when we plug in 2 for x, we get a value for the function of 0. That means we can put a y in the rightmost column and, lucky for us, stop working. We found our first

Now that we've found one zero, which we know is now at x equal 2, let's see if we can find some more. In order to do that, we're going to need to divide this polynomial by some factor that's associated with this route we just found. So please tell me then, what linear factor produces a zero at x equals 2? Type your answer in this box.

The factor we're looking for is x minus 2, since we know that if we set this equal to 0, that would lead us to x equaling 2. That means this is the factor we're going to need to divide this polynomial by.

Let's do that division then. Please use polynomial long division to divide x cubed minus 20x squared plus 118x minus 164 by x minus 2. Write your answer in this lovely green box.

The answer we get when we divide is x squared minus 18x plus 82. And sure enough, we have a remainder of 0, showing that 2 really is a zero of this function.

So we've already factored out an x minus 2 from our original polynomial, which leaves us with this quadratic expression, x squared minus 18 x plus 82. Now to find the last two factors of this cubic function we started out with, we need to find the factors of what we have left over, this new second degree expression. Now, it doesn't look like we can just factor this. Considering I don't know anything that multiplies to 82, but adds to negative 18. So, I think we should use a quadratic formula. Remember that if we label the coefficients of a quadratic expression, a, b and c, then we can find zeros of that expression using this formula. I'll just leave it down here so you can use it in the problem. So please do exactly what we were just discussing. Use the quadratic formula to find the roots of x squared minus 18 x plus 82. Enter one of them in this box and one of them in this box.

Going through all the steps of the quadratic formula gives us 9, plus or minus i. Or, in other words, our zeroes are 9 plus i and 9 minus i. Interesting. So I guess that cubic functions, like the one we started out with, can have imaginary zeros too. That's pretty cool. Think for a second about what this means for the graph of Grant's profit function.

Now Athena is well equipped to answer Grant's question. How many break even points are there? Remember, a break even point is just a point on the graph, where profit or P of x equals 0. Please fill in the number of those spots right here. Then, please tell me what the value of x is at each of these break even points. Separate your answers by commas.

We can figure this out by looking at the roots, or the zeros of this function. We found before that they were 2, 9 plus i and 9 minus i. Now, these breakeven points are going to be actual points on the graph. And remember, since our graph is only plotting real values of x and real values of y, 9 plus i and 9 minus i, are not actual x values on the graph. That means the only real x intercept is at x equals 2. So there's only one break even point, and as I just said the value of x at that point is 2. Now remember, x in this equation, stood for thousands of dollars spent on advertising. So Grant doesn't just need to spend $2 to break even, he needs to spend $2,000. He'd probably rather spend two but unfortunately, advertising is pretty expensive.

Athena and Grant have shown us that all sorts of polynomials, not just quadratics, can have imaginary zeroes. So now we found zeros that are integers, rational numbers, approximated numbers, and even zeros to the imaginary parts. I think it would be a good idea if at this point we took a step back, and talked about how all these different kinds of zeros can work together to build up a single polynomial function. Remember, a polynomials zeros, factors in degree, are all very closely related to one another.

Here are two functions and the graphs that go along with each of them. For each of these graphs, I would like you to fill in the proper numbers to complete the sentence that goes along with it. In other words please tell me for each one how many linear factors and how many zeros it has and what degree it is

Function A, f of x equals x minus 2 times x minus 1, has two linear factors, which we can see right here, two zeros, which we see with these 2 x intercepts, and is degree 2. Interesting. Function B, which is f of x equals x minus 2 times x minus 1 times x plus 3, has three linear factors, three zeros, and is degree 3. Now, please note that, even though I pointed out the x intercepts on these graphs when we talked about zeroes, if the zeroes were non real, then we wouldn't see them as x intercepts. But they would still be roots.

I'm thinking of a function of degree four. How many linear factors does my function have? Remember a linear factor is something of the form x minus n, where n can be any complex number. Remember complex numbers can be either real or imaginary or a combination of real and imaginary.

The answer is 4. Let's think about this for a second. Let's say, for a moment, that our function had five linear factors, then we would have five linear factors. Five factors of x minus some number all multiplied together. Now, when we expand this multiplication, in other words, when we distribute all of these factors to one another, we're going to end up with some term that has an x to the fifth power. That's no good. This is a fifth degree polynomial, not a fourth degree. Considering this, you know that one of these factors have to go away, in order for us to only have the polynomial of degree 4.

Here's another function for us to consider. This time, we have f of x equals x times x minus 2 times x minus 2 times x minus 1 times x minus 1. Well, this is interesting. We have some repeated factors here. So far, we've only seen polynomial functions. That have unique zeros or all factors with different zeros associated with them. In the past when we were looking at these functions with unique zeros the number of zeros was always equal to the degree of the polynomial. I wonder if that's the case if we have some zeros that are repeated instead? Let's use this function to check that out. What I'd like you to do to analyse this function then is to enter the x value of each of its zeros and then tell me how many times each of those factors associated with those zeros occur in the equation for the function. So if you said that one of the zeros was four for example, and you saw the factor x minus 4 three times in this equation, you would write a three right here. Lastly, I'd like you to add up the total number of zeros that we have. This total number should just be the sum of all of the numbers in this repeat boxes. Think, then, about how this should relate to the total number of factors we see in this equation.

The first factor we see is just X, which is associated with a 0, X equals 0. There's only one of them so I write a 1 in this box. The other two 0's find are at X equals 1 and X equals 2. Each of the factors associated with these appears twice in the equation. If we add these numbers up we see that there are five expected, the total number of zeros we have 5 is equal to the number of factors or rather linear factors we see in the polynomial, and is also equal to the degree of this polynomial function. Once again we see that the linear factors zeros, in the degree of a polynomial function, are all very closely connected.

Zeros can be any kind of complex number. Some of them are just real numbers, but some of them are complex numbers that have imaginary parts. Let us look at a little bit more closely at the case where our zeros are not just real numbers. When we looked at a quadratic equations, we found that if we had a solution of x equals 2 plus 3 i, then it's complex conjugate was also a solution. So, can you tell me what the complex conjugate of 2 plus 3 i is?

The complex conjugate of 2 plus 3i is just 2 minus 3i. That means that if we know that 2 plus 3i is one of our roots, we also automatically know that 2 minus 3i is one of our roots.

Now, if we found that 2 plus 3i and 2 minus 3i are both roots of some polynomial, then we also already know two factors of this polynomial. Those factors are x minus the quantity 2 plus 3i and x minus the quantity 2 minus 3i. If we know that both of these are factors of some polynomial, then they're going to be multiplied together to help create it's equation. So, what's an equation of a polynomial that only has these two factors? I've written them out as multiplied for you right here. But, I'd like you to expand this multiplication and simplify, writing the final version of this polynomial in this blue box.

Now, the first way I thought of solving this problem, or just simplifying the left hand side rather, was to go through and multiply every term from the first set of parentheses by every term in this second set of the brackets. But that way, we end up this really, really long polynomial, and it does reduce, but it's not very pretty. So, I'm going to show you a different way of thinking about this multiplication. We can start off by distributing this negative sign to each of the terms inside this inner set of parentheses. We can do this for both of the sets of brackets. Now, this is pretty interesting. You see that we have x minus 2 here, and x minus 2 here. Why if we see if we can just consider those as a unit, and see what happens. At this point, we can recognize that we're going to end up with a difference of two squares as our final polynomial. Let's see that happen. Since we have an x minus 2 in both of these expressions, and then opposite signs of negative 3i and positive 3i here, we end up with x minus 2 squared plus 9. Notice I've already multiplied the i's together and switched the sign. Multiplying out this square and adding the 9 gives us a final polynomial of x squared minus 4x plus 13 equals 0. And this is a pretty cool trick, definitely saved a lot of space and a lot of writing for me. And, you know, I always love ways to be a little bit more lazy with my math.

I was using my favorite red pen to write with yesterday, but unfortunately it had an accident. It exploded all over my piece of paper, and you can see I ended up with a pretty big mess. I even got it all over my hands. I don't know if you can tell but my fingers were pretty red yesterday. Now, the big problem however, is that I lost a bunch of the work I had done. I started out with a polynomial and then found all of its zeros. But now I can only see three of them. Think you can make it out here. 1, 2 and i. So I'm trying to figure out how much work I need to do to figure out my last zero. Do I need to actually recalculate it? To answer this question, please just write yes or no in this box. In the event that you write no in this box, and we can find this fourth zero without really doing any math at all, write what the fourth zero is in this box.

And the answer is no. We actually don't need to do any calculations or no major calculations, at least, to find the last zero. It's simply the complex conjugate of this one non-real zero we already have, i. The complex conjugate of i is negative i. So that is our missing zero. Awesome.

Let's say we have some other 4th degree polynomial? And we already know two of it's roots. They are 2 minus 3i and 4 plus 5i. What are the two 0's that we're missing?

The two zeros were missing is just the complex conjugates of the roots that we already know. So we'll have 2 plus 3i, and 4 minus 5i.

I'm really enjoying this practice of creating functions when we already know about their zeroes. So, let's keep doing that for a bit. Let's say that, this time, we have a polynomial that only has zeroes of 2, 3, and 4. And none of these are repeated. There's just one factor coming from each of these. What then, is the degree of that polynomial?

It's a polynomial of degree 3.

Let's keep working with this same polynomial, with these same zeros. How many factors does a polynomial have?

It has three factors. This is perfectly in line with the fact that it's also a degree three.

Now that we know that there are three factors, what are these factors? Please write them in the form x minus a, and put one in each of these boxes.

Our linear factors will be x minus two, x minus three and x minus four.

So, what's the simplest polynomial function that we can create from the three factors we just found? Note that the only factors this polynomial function will have are these three, it won't have any others. You don't need to expand or simplify the expression, just write it in factored form right here.

The polynomial we're looking for is simply the product of all three of these factors, each put inside parentheses, of course. So we have f of x equals x minus 2, times x minus 3, times x minus 4. Awesome.

So now, we found one polynomial function that has these linear factors as it's only linear factors. But is this the only function with just these linear factors? Please answer yes or no, think carefully about this.

And the answer is no. Even without multiplying in another linear factor, we can multiply other numbers in. For example, if I multiplied in a factor of 3, we would still only have zeroes at 2, 3, and 4. Multiplying in a constant term, doesn't change the linear factors.

Let's think about what we've learned. Now, we know that a function has the same number of zeros, and the some number of linear factors as its degree. And, of course, we need to make sure that we count the repeats here. We've also seen that complex zeros always come in pairs as conjugates. And we know how to find zeros of all different types. I think it's time then, to work through an example that involves multiple techniques we've learned. This is going to be a great problem to help you summarize all the important aspects of determining zeros. So, get ready for this amazing polynomial function. We have f of x equals, we're off to a pretty good start. 2x to the fourth, minus 13x cubed, plus 32x squared, minus 13x, and that's it. I hope you're impressed. So, if we want to find the zeros of this function, how do you think the best way to start would be? Should we factor the first two terms and then factor the last two terms? Should we grab a snack, since it's going to be a long ride? Should we pull out the greatest common factor from all of the terms? Should we use the rational zero theorem involving negative 13 and 2? Should we divide both sides by x? Or should we combine like terms?

We should pull out the greatest common factor from all the terms here. That greatest common factor we can see is x.

Now that we've recognized that there's a common factor of x among all these terms, let's pull it out to the front. When we do that, what belong's inside the parentheses here?

To figure out what each of the terms should become inside the parentheses, we just need to divide each of them by x, since we've written our factor up here. Now our function reads f of x equals x times the quantity 2x cubed minus 13x squared plus 32x minus 13.

We've already found one of the factors of our original polynomial. So, what should be our next step in continuing this factoring process? Should we factor by grouping for the polynomial inside the parenthesis? In other words, should be factor the first two terms and then the second two terms? Should we distribute multiplication by x to all the terms inside the parenthesis? Should we combine like terms? Should we divide both sides of the equation by X? Or should we find factors of negative 13 and 2, and then apply the rational zero theorem?

The last choice is the best one. Factoring by grouping won't work for this cubic. We just pulled out next, so we don't want to remultiply it back in. There are no like terms to combine. And if we divide both sides by x, we'll end up with f of x over x here. And that's not really very helpful. The best way to find potential zeroes, of this polynomial, is to use the rational zero theorem.

Our solution then is to find the factors of 2 and negative 13, so that we can eventually use the rational zero theorem. So, what are the positive factors of

The positive factors of 2 are just 1 and 2, and the positive factors of negative 13 are just 1 and 13.

Now that we found the factors of our constant term and our leading coefficient for this polynomial inside parentheses, let's use these in the rational zero theorem. Please write out for me, all of the zeros that the rational zero theorem says this polynomial might have. Separate your answers by commas and don't forget the negative versions of your roots.

To find all these potential zeroes, we know that, for each one, we're going to take a factor of negative 13, and divide it by a factor of 2. Then we need to the positive and negative versions of each of these ratios. So we could have 1 over 1, which is 1. Or the negative version of that, negative 1. 13 over 1, which is just 13. And negative 13, as well. In the end, we come up with eight roots, so the rational of zero theorem tells us this function might have.

Now that the rational zero test has given us eight possibilities for zeroes at this function. Why don't we test some of them out? Now, of course, figuring out the value of the function at any of these values for x will give us useful information. But right now, I'm really only interested in which ones of these, when plugged into our function, give us f of x 0. Or rather, f of whatever value we plug in equals 0. Here's how we're going to go about doing this. I've made a list of the values of the function at all the possible zeroes we had, and I'd like you to evaluate all of these function values. All you need to do for this quiz is, starting at the top of the list, work your way down, and as soon as you find a value of the function that equals 0, write a 0 in the box next to that value. You can leave all the other boxes blank, and remember, you can stop after you have found just one zero. The top most one.

If we work our way down through this list, we find that the first zero value is when x equals 1 half. f of 1 half equals 0.

Now that we've found out another 0 of our function, we should automatically know what one of its factors is. One of its linear factors to be precise. So, what factor is associated a 0 of 1 half. Please fill in just integers in these two boxes, and make sure that the numbers you fill in have a greatest common factor of 1.

Now you might have been tempted to write that our linear factor was x minus one half, since if we set this equal to 0 and solve for x we get one half. However I want an integer coefficient in constant terms here. This way we can transform our linear factor to be like that. Is by multiplying both sides by 2, since that's the denominator of our fraction here. If we do that, we end up with a linear factor instead of 2x minus 1. We can double check, and sure enough, the greatest common factor between 2 and negative 1, is 1. Perfect, this must be one of our linear factors.

We are one step closer now, to getting this function in its fully factored form. Now, we know two of its linear factors. We already found x. And now, we also know that 2x minus 1 is a linear factor. The next thing for us to do then, is figure out what's left over if we pull out a factor of 2x minus 1 from this part of the expression. So, how do we do that? Well, we need to divide this expression inside the parentheses by 2x minus 1. Please use long divinsion to divide 2x cubed minus 13x squared plus 32x minus 13 by 2x minus 1. What do we get? Write your answer in this green box.

Our long division here works here just the same as it always has. When we go through all the steps, we end up with x squared minus 6x plus 13. And luckily, we have a remainder of 0. Which as always, shows that 2x minus 1, really is a factor of the expression we started out with. Since the remainder is 0, that means if we plug in the root associated with this, the function will have a value of 0.

Let's take a step back, and see how far we've come in solving this problem. We started out with our function, f of x equals 2 x to the fourth, minus 13 x cubed, plus 32 x squared, minus 13 x. And so far, we've pulled out two linear factors. Now we have x times 2 x minus 1, times x squared, minus 6 x plus 13. We are clearly well on our way to having this fully factored. All we have left to do, is figure out the zeros that go along with this quadratic expressions right here. What then, are the zeros of x squared minus 6 x plus 13? Please write them in these boxes.

If we set this expression equal to 0, I can't really think of a way to factor it. So, our best way is to use the quadratic formula. All we need to do is plug the proper coefficients into the proper places in the formula, and then simplify. And it looks that we're going to get an answer that has an imaginary part since we're taking this square root of a negative number. We end up with 3 plus or minus 2i which, then of course, we can split into two separate answers. With 3 plus 2i and 3 minus 2i.

We just found the last two zeroes of our function, f of x. They are 3 plus 2i and 3 minus 2i. To at long last get our function in fully factored form, we just need to figure out what linear factors correspond to these two zeroes. Please write 1 in each of these boxes.

One of our factors will be x minus the quantity 3 plus 2i and the other will be x minus the quantity 3 minus 2i. We can, of course, distribute the subtraction if we want to. If we do that, then we'll end up with x minus 3 minus 2i and x minus 3 plus 2i. Either way of writing them works.

Now finally, since we have these last two linear factors, we can write f of x as the product of four linear factors, which we knew it would have because it is a function of degree 4. So now we can write that f of x equals x times 2 x minus 1 times x minus 3 minus 2 i times x minus 3 plus 2 i. This is really awesome. We've made so much progress since we started off. Now I have one final question for you, and it's really what we've been working toward for the entire problem. What are all the zeros of f of x? We already know two of them from a couple of quizzes ago, 3 plus 2 i, and 3 minus 2 i. And so I'd just like you to fill in the last two. Check in with the linear factors we haven't used yet to find zeros, but at least haven't used in this quiz yet.

The zero that goes along with the linear factor of x is just 0, and the one that goes along with 2x minus 1, is what we found before, one half. Look at that, amazing. We found all four zeros of our wonderful polynomial. And along the way, we've worked on a ton of different techniques we've been learning over the past few lessons. Awesome job.

By now, you've learned about many different tools that we can use to get polynomials into their full factored forms. Sometimes, we can just pull a common factor out of all the terms in an equation. Other times, we need to use the rational zero theorem in conjunction with polynomial long division. In the event that we have a second degree polynomial, we can use the quadratic formula. All of these methods tell us about exact values of roots. But in the event that we're having trouble finding these exact values, we can always use the intermediate value theorem to approximate zeroes.