WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about the intermediate value theorem and polynomial division.
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Previously, we talked about how we need to factor a polynomial to find its roots.
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But we recently saw the quadratic formula, which gave us the root of any quadratic without even having to factor at all.
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So, maybe we don't need factoring; maybe there are formulas that allow us to find the roots for any polynomial; wouldn't that be great?
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We would just be able to plug things in, put out some arithmetic, and we would have answers, no matter what polynomial we were dealing with.
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Not really--while there are root formulas for polynomials of degree 3 and 4, they are so long and complicated, we are not even going to look at them.
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The formula for finding the roots of a cubic, of a degree 3 polynomial, is really long and really complex.
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And it is just something that we don't really want to look at right now.
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And a degree 4 would be even worse; so we are just not going to worry about them.
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And then, not only that--they just simply don't exist for degree 5 or higher.
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So, if you are looking at a degree 5 or higher, there is no such formula for any degree 5 or higher thing.
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It was proven in 1824 that no such formula can exist, that would be able to do that.
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Thus, it looks like we are stuck factoring, if you want to find the precise roots of a higher-degree polynomial.
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If we are working with a higher-degree polynomial, and we need to know its roots for some reason, we have to figure out a way to factor it.
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In this lesson, we are going to learn some methods to help factor these complicated polynomials.
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We will first learn a theorem to help us guess where roots are located, and then a technique for helping us break apart big polynomials.
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All right, let's go: In the lesson Roots of Polynomials, we mentioned the theorem that every root implies a factor.
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That is, if f(x) is a polynomial, then if we have f(k) = 0 (k is a root), then that means we know x - k is a factor of f(x),
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because if k = 0, then that means that x = k causes a root; so x - k = 0; and thus, we have a factor from our normal factor breakdown.
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For example, if we have g(x) = x³ - 3x² - 4x + 12,
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and we happen to realize that when we plug in a 2, that all turns into a 0
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(and it does), then we would know that g(x) is equal to (x - 2), this thing becoming (x - 2), times (_x² + _x + _).
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We know that there is some way to factor that polynomial where something is going to go in those blanks.
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2 is a root; this theorem tells us that (x - 2) must be a factor.
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It doesn't tell us what will be left; but it does make the polynomial one step easier.
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We know we can pull up (x - 2), so then we can use some logic, play some games, and figure out what has to go in those blanks.
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But notice: it doesn't directly tell us what is going to be there.
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If we are lucky, we can sometimes find a root or two purely by guessing.
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We might think, "Well, I don't know where the roots are; but let's try -5; let's try √2; let's try π."
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We might just try something, and surprisingly, it ends up working; that is great.
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But is this always going to end up being the case?
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If we manage to pick something where we figure out the root, and then we figure out something that gets us a root;
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we plug in a number, and we get 0, and we know we have a root.
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And if we have a root, that means we have a factor.
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Knowing a factor makes it that much easier to factor the whole polynomial.
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But it is hard to guess correctly every time; guessing is guessing--you can't guess every single time.
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Luckily, there is a theorem that will give us a better idea of where the roots are located.
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The intermediate value theorem will help us find roots; it goes like this:
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If, first, f(x) is a polynomial (for example, we have this nice red curve here; that is our f(x));
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then a and b are real numbers, such that a < b.
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What that means is just that a and b are going from left to right; a is on the left side, and then we make it up to b.
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That is what this a < b is--just that we know an order that we are going in.
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And then, u is a real number, such that f(a) < u < f(b).
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So, we look and figure out that part; we see that, at a, we are at some height f(a); and at b, we are at some height f(b).
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That is how we get that graph in the first place.
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Then, u is just something between those two heights; so u is just some height level,
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where we put an imaginary horizontal line that ends up saying,
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"Here is an intermediate value between f(a) and f(b), some intermediate height between those two."
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The intermediate value theorem tells us that there exists some c contained in ab, such that f(c) = u.
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So, we are guaranteed the existence of some c that is going to end up giving us this height, u.
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Basically, if we have some height u, and that height u crosses between two different heights
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that go...we have two points going from left to right, so we are going from left to right, and we cross over some height during that thing;
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we start lower, and then we end above; we are guaranteed that we had to actually cross it.
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We had to go across it; and since we had to go across it, there must be some c where we do that crossing--where we end up landing on that height.
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There is something that will give us that intermediate value.
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Why does this have to be true? Because polynomials are continuous; there are no breaks in their graphs.
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The only way f could possibly manage to not end up being on this height u--the only way f could dodge the height u-- is by jumping this intermediate height.
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The only way that we have this...the graph is going; the graph is going; the graph is going; the graph is going; the graph is going.
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And then, all of a sudden, it would have to jump over that height to be able to manage to not end up touching it.
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If our graph touches the height at any point, then we have whatever point is directly below where it touched that height;
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that is the location that intersects that intermediate value.
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That is the thing that is going to fulfill our intermediate value theorem.
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So, on any polynomial or any continuous function (in fact, the intermediate value theorem is true for any continuous function,
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but we are just focusing on polynomials), they can't jump; polynomials--continuous functions--they are not allowed to jump.
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There are no breaks in their graphs; so since there are no breaks, they have to end up crossing over this height.
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Since they cross over this height, there is some place on the graph...we just look directly below that,
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and that guarantees us our c, where f(c) is going to be equal to u; there we go.
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This means we can use the intermediate value theorem to help us find roots.
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If we know that f(a) and f(b) are opposite signs--so, for example, if we know that at a, f(a) is positive--
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we have a positive for f(a)--and then, at b, we know that we are negative--we have a negative for f(b)--
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then we know that there has to be a root.
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Why? Well, we have to have some f that is going to make it from here somehow to here.
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It has to manage to get both of those things.
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So, the only way it can do it is by crossing over at some location.
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It might cross over multiple times, but it has to cross over somewhere.
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Otherwise, it is not going to be able to make it to that point that we know is below y = 0.
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Since f must cross over y = 0, we are guaranteed the existence of this c at some point where it ends up crossing over.
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Now, this does not mean that there is only one root; like we just saw in that second thing I drew, it could cross over multiple times.
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This theorem guarantees the existence of at least one root, but there could be multiple roots.
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if it bounces back and forth over that y = 0 on the way to making it to the second point.
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OK, so say we find a root of a polynomial by a combination of luck and the intermediate value theorem.
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We somehow manage to figure them out, or the problem just tells us a root directly from the beginning.
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In either case, with the root, we now know a factor; a root tells us a factor.
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But how can we actually break up the polynomial if we know a factor?
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How do we divide a polynomial by a factor?
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For example, say we know x = 3 is a root of this polynomial.
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Then we know that there is some way to divide out x - 3 so that we have x - 3 pulled out,
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and some _x³ + _x² + _x + _--
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there is some other polynomial that is going to go with it, that the two will multiply.
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Otherwise, it would not have divided out cleanly.
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It couldn't be a factor unless there was going to be this other polynomial where it does divide out cleanly.
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So, how do we actually find out that thing that happens after we divide out this polynomial?
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How do we do polynomial division?
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To explore this idea, let's refresh ourselves on long division from when we were young.
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So, long ago, in grade school and primary school, we were used to doing problems like 1456 divided by 3.
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Let's break out long division: we have 1456, so the first thing we do is see how many times 3 goes into 1.
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3 goes into 1 0 times; so it is 0 times 3; that gets us 0 down here; we subtract by 0; nothing interesting happens yet.
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1; and then we bring down the 4; so we have 14 now.
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How many times does 3 go into 14? It goes in 4 times.
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3 times 4 gets us 12; we subtract by 12: minus 12; so minus 12...we get 2.
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Then, we bring down the next one in the running; let's keep the colors consistent.
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We have 5 coming down, so we now have 25; how many times does 3 go into 25?
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It goes in 8 times...16, 24...so now we subtract by 24; 25 - 24 gets us 1.
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We bring down the 6; we have 16; how many times does 3 go into 16? It goes in 5; we get 15.
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So, minus 15...we get 1; now, we don't have any more numbers to go here.
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There is nothing else, so that means we are left with a remainder of 1.
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We have whatever our very last thing was, once we ran out of stuff; that becomes our remainder: 485 with a remainder of 1.
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If we wanted to express this, we could say 1456 divided by 3; another way of thinking of that is 1456 is equal to 3(485) + 1.
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Or, alternately, if we wanted to, we could say 1456/3 is equal to 485 plus the remainder, also divided,
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because we know that we get the 485 cleanly, but the 1 is a remainder.
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So, it doesn't come out cleanly; so it comes out as 1/3.
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You could also see the connection between these two things, because we simply divide both sides by 3;
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and that is how we are getting from one place to the other place.
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That is what we are getting by going through long division.
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Really quickly, let's also look at this 1456 = 3(485) + 1; we call this right here the **dividend**; the thing being divided is the dividend.
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This one here is the thing doing the dividing; we call the thing doing the dividing the **divisor**.
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Then, what we get out of it is our quotient; what comes out of division is the **quotient**.
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And finally, what we have left at the very end is the **remainder**.
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All right, these are some special terms; you might not remember those from grade school.
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But these are the terms that we use to talk about it.
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Why does that matter? because we are now going to want to be able to express it in a more abstract, interesting way,
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where we are talking, not just about real numbers, but being able to talk about polynomials.
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We found that 1456 divided by 3 became 3 times 485 plus 1.
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Clearly, we can do this method for any two numbers; and it turns out that we can do a very similar idea for polynomials.
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We call this the division algorithm--this idea of being able to do this.
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And it says that if f(x) and d(x) are both polynomials, and the degree of f is greater than or equal to d--
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that is to say, f is a bigger polynomial than d--and d(x) is not equal to 0 (why does d(x) not equal 0?
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because we are not allowed to divide by 0, so if d(x) is just simply 0 all the time, forever,
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then we can't divide by it, because we are not allowed to divide by 0)--given these things (f(x) and d(x), both polynomials;
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f(x) is a bigger polynomial--that is to say, higher degree than d--and d(x) is not simply 0 everywhere),
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then there exist polynomials q(x) and r(x) such that f(x) = d(x) times q(x) plus r(x).
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So, how is this parallel? f(x) is the thing being divided.
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The thing being divided is our dividend, once again.
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The thing doing the dividing is the divisor.
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What results after we have done that division is the quotient.
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And finally, what is left at the end is our remainder.
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So, the remainder is right here, and our quotient is right here.
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So, we have parallels in this idea of 1456/3; we have this same thing coming up here.
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1456/3 is not equal to 3(485); it becomes this idea; so it is not actually equal: 1456/3 is 485 plus 1/3; but it becomes this idea.
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So, the dividend here, the thing that we are breaking up, in this idea, is 1456; let's just knock this out, so we don't get confused by it.
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Our divisor, the thing doing the dividing, is 3; what we get in the end, our quotient, is 485; and the remainder is that 1; 1 is left out of it.
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Now, we could also have an alternative form where we write this as f(x)/d(x) = [q(x) + r(x)]/d(x),
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where we just turn this into dividing; we get between these two by dividing by...not 3...but dividing by d(x), dividing by our polynomial.
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So, basically the same thing is happening over here, so this would not be 3 times 45; it should be 1/3.
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1456/3 is equal to 485 + 1/3, because that is 1456/3.
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We have a real connection between these two things.
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The division algorithm is giving us this idea that if we have some polynomial f(x),
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we can break it into the divisor, times the quotient, plus some remainder, which, alternatively, we can express as
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the polynomial that we are dividing, divided by its divisor, is equal to the quotient, plus the remainder, also divided by the divisor.
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This is effectively a way of looking at f(x) dividing d(x)--seeing what is happening here.
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Now, notice: r(x) is the remainder, so in the case when r(x) = 0, then that means we have no remainder,
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which we describe as d(x) dividing evenly; so when it divides evenly into f(x), then that means d(x) is simply a factor.
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5 divides evenly into 15, so that means 5 is a factor of 15.
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How do we actually use the division algorithm to break apart a polynomial?
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Let's look at two methods: we will first look at long division, and then we will look at synthetic division.
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First, long division: we will just take a quick run at how we actually use polynomial long division.
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And it works a lot like long division that we are already used to.
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So, let's see it in action first; and then we will talk about how it just worked.
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We have x⁴ - 5x³ - 7x² + 29x + 30, divided by x - 3.
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So, we are dividing x - 3; it is dividing that polynomial: x⁴ - 5x³ - 7x² + 29x + 30.
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OK, the first thing we do is ask, "All right, how many times does x - 3 go into x⁴ - 5x³?"
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Well, really, we are just concerned with the front part; so just look at the first term, x.
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How many times does x go into x⁴?
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Well, x⁴ divided by x would be x³; so the x³ goes here.
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Now, that part might be a little confusing: why didn't we end up having it go at the front?
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Well, think of it like this: if we have 12 divide into 24, does 2 show up at the front?
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No, 2 doesn't show up at the front; 2 shows up on the side, because 12 is 2 digits long; so we end up being at the second-place digit, as well.
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It is 2 digits long, so we go with the second-place digit.
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So, the same thing is going on over here: x - 3 is two terms long, so we end up being at the second term, as well.
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All right, so all of the ideas...we are going to knock them out really quickly, now that we have them explained.
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So, that is why we are not starting at the very first place--because we have to start out at where they line up appropriately.
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We check first term to first term, but then we go as far wide as that thing dividing is.
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So, x³ is what we get out of x⁴ divided by x.
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So, now we take x³, and we multiply (x - 3), just as we did in long division.
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x³ times (x - 3) becomes x⁴ - 3x³.
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Now, we also, in long division, subtracted now; so subtract.
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Let's put that subtraction over it; minus, minus, 2 minus's become a plus...so we have -x⁴ attacking x⁴, so we have 0 here.
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And 3x³ + -5x³ becomes -2x³.
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And then, the next thing we do is bring down the -7x².
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So, -7x²: now we ask ourselves, "How many times does x go into -2x³?"
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Well, that is going to go in -2x², so we get -2x².
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-2x² times x - 3 becomes -2x³; -2x² times -3 becomes + 6x².
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Now, we subtract by all this stuff; we distribute that; that becomes positive; this becomes negative.
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We now add these things together, so -2x³ + 2x³ becomes 0 once again.
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-7x² - 6x² becomes -13x².
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The next thing we do is bring down the 29x, so + 29x.
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How many times does x go into -13x²? That goes in -13x.
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The -13x times x - 3 gets us -13x² + 39x; that is this whole quantity; subtracting that whole thing,
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we distribute that, and it becomes addition there, and subtraction there.
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So, we have -13x² + 13x²; that becomes 0 once again; 29x - 39x becomes -10x.
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Once again, we bring down the 30; so we have + 30 here.
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And now, how many times does x go into -10x? x goes in -10 times.
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So, -10 times x - 3 is -10 + 30; we subtract this whole thing; we distribute that; and we get 0 and 0.
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So, we end up having a remainder of 0, which is to say it goes in evenly.
00:20:19.300 --> 00:20:29.800
So, if that is the case, we now have x³ - 2x² - 13x - 10 as what is left over after we divide out x - 3.
00:20:29.800 --> 00:20:36.200
So, we know our original x⁴ - 5x³ - 7x² + 29x + 30 factors as...
00:20:36.200 --> 00:20:48.400
let's write this in blue, just so we don't get it confused...(x - 3)(x³ - 2x² - 13x - 10).
00:20:48.400 --> 00:20:52.000
That is what we have gotten out of it; cool.
00:20:52.000 --> 00:20:55.500
To help us understand how that worked, let's look at the steps one at a time.
00:20:55.500 --> 00:21:00.200
You begin by dividing the first term in the dividend by the first term in the divisor.
00:21:00.200 --> 00:21:05.700
So, our dividend is this thing right here; its first term is x⁴; the first term of our divisor,
00:21:05.700 --> 00:21:09.400
the thing doing the dividing, is x; so how many times does x go into x⁴?
00:21:09.400 --> 00:21:19.500
Well, x⁴ divided by x...if we are confused by the exponents, we have x times x times x times x, over x;
00:21:19.500 --> 00:21:27.100
so, one pair of them knock each other out; so we have x³ now, x times x times x; great.
00:21:27.100 --> 00:21:30.900
That is why the x³ goes here; it is the very first thing that happens.
00:21:30.900 --> 00:21:38.300
The next thing: we take x³, and we multiply it onto (x - 3).
00:21:38.300 --> 00:21:44.000
So, x³(x - 3) becomes x⁴ - 3x³.
00:21:44.000 --> 00:21:50.300
You multiply the entire divisor (this right here) by the result, our x³.
00:21:50.300 --> 00:21:54.800
And then, we subtract what we just had from the dividend.
00:21:54.800 --> 00:21:58.900
x⁴ - 3x³: we subtract that from x⁴ - 5x³.
00:21:58.900 --> 00:22:04.900
This gets distributed, so we get a negative here, a plus here; and so that becomes -2x³.
00:22:04.900 --> 00:22:11.500
The next thing we do is bring down the next term; so our next term to deal with is this 7x².
00:22:11.500 --> 00:22:17.500
It gets brought down, and we have -2x³ - 7x².
00:22:17.500 --> 00:22:23.100
And then, once again, we do the same thing: how many times does x go into -2x³?
00:22:23.100 --> 00:22:34.300
It goes in -2x²; so then, it is -2x²(x - 3); and we get -2x³ + 6x².
00:22:34.300 --> 00:22:38.100
We subtract that, and we keep doing this process until we are finally at the end.
00:22:38.100 --> 00:22:42.400
We might have a remainder if it doesn't come out to be 0 after the very last step.
00:22:42.400 --> 00:22:45.500
Or if it comes out to be 0, we are good; we don't have a remainder.
00:22:45.500 --> 00:22:52.200
All right, there is also a shortcut method that goes a bit faster if the divisor is in the form x - k.
00:22:52.200 --> 00:22:58.000
And notice: it has to be in the form x - k; if it is in a different form, like x² + something, we can't do it.
00:22:58.000 --> 00:23:05.700
Now, notice that you could deal with x + 3; it would just mean that k is equal to -3, so that is OK.
00:23:05.700 --> 00:23:10.800
It just needs to be x, and then a constant; so that is the important thing if we are going to use synthetic division.
00:23:10.800 --> 00:23:15.900
So, it goes like this: we let a, b, c, d, e be the coefficients of the polynomial being divided.
00:23:15.900 --> 00:23:21.000
For example, if we have ax⁴ + bx³ + cx² + dx + e,
00:23:21.000 --> 00:23:27.300
then we set it up as follows; this k right here is on the outside of our little bracket thing.
00:23:27.300 --> 00:23:30.100
And then, we set them up: a, b, c, d, e.
00:23:30.100 --> 00:23:37.100
Now, the very first step: every vertical arrow--you bring whatever is above down below the line.
00:23:37.100 --> 00:23:43.100
So, a, since there is nothing underneath it...we add terms on vertical arrows, so they come down adding together.
00:23:43.100 --> 00:23:46.100
So, a comes down; there is nothing below it, so it becomes just a.
00:23:46.100 --> 00:23:55.100
Then, you multiply by k on the diagonal arrow; so we have a; it comes up; we multiply by k, and so we get k times a.
00:23:55.100 --> 00:23:58.900
Then, once again, we are doing another vertical arrow where we are adding.
00:23:58.900 --> 00:24:06.100
We go down: k times a...b + ka becomes ka + b.
00:24:06.100 --> 00:24:12.300
The next thing that will happen (you will probably want to simplify it, just to make it easier, but) we multiply that whole thing by k once again.
00:24:12.300 --> 00:24:15.800
And we keep up the process until we get to the very last thing.
00:24:15.800 --> 00:24:19.300
And the very last thing is our remainder.
00:24:19.300 --> 00:24:28.500
All the terms preceding that, all of the terms in these green circles, are the coefficients of the quotient.
00:24:28.500 --> 00:24:33.500
So, if this is one, then we will have a constant here, starting from the right; and this will be x's coefficient;
00:24:33.500 --> 00:24:36.500
this will be x²'s coefficient; this will be x³'s coefficient.
00:24:36.500 --> 00:24:38.800
And that makes sense: since we started with that to the fourth,
00:24:38.800 --> 00:24:43.100
and we were dividing by something in degree 1, we should be left with something of degree 3.
00:24:43.100 --> 00:24:45.500
All right, let's see it in action now.
00:24:45.500 --> 00:24:51.400
Once again, dividing the same thing, we have x⁴ - 5x³ - 7x² + 29x + 30.
00:24:51.400 --> 00:25:00.200
So, we have x - 3; remember, it is x - k, so that means our k is equal to 3, because it is already doing the subtraction.
00:25:00.200 --> 00:25:11.500
We have 3; and we set this up; our first coefficient here is just a 1; 1 goes here.
00:25:11.500 --> 00:25:16.100
What is our next coefficient? -5; -5 goes here.
00:25:16.100 --> 00:25:19.400
What is our next coefficient? -7; -7 goes here.
00:25:19.400 --> 00:25:26.300
Our next coefficient is 29; what is our next coefficient? 30, and that is our last one, because we just hit the constant.
00:25:26.300 --> 00:25:32.900
All right, so on the vertical parts, we add; so 1 + _ (underneath it) becomes 1.
00:25:32.900 --> 00:25:46.600
Then, 3 times 1 becomes 3; -5 plus 3 becomes -2; -2 times 3 is -6; -6 plus -7 becomes -13.
00:25:46.600 --> 00:25:57.900
3 times -13 becomes -39; 29 plus -39 becomes -10; 3 times -10 becomes -30; 30 + -30 becomes 0.
00:25:57.900 --> 00:26:04.300
Now, remember: this very last one is our remainder; so our remainder is 0, so it went in evenly, which is great,
00:26:04.300 --> 00:26:10.800
because since we just did this with polynomial long division, and we saw it went in evenly, it had better go in evenly here, as well.
00:26:10.800 --> 00:26:18.100
So, this is our constant right here (working from the right); this is our x; this our x²; this is our x³.
00:26:18.100 --> 00:26:27.100
So, we get x³ - 2x² - 13x - 10; that is what is remaining.
00:26:27.100 --> 00:26:38.100
So, we could multiply that by x - 3; and then this whole expression here would be exactly what we started with in here, before we did the division.
00:26:38.100 --> 00:26:42.300
Great; all right, so which of these two methods should we use?
00:26:42.300 --> 00:26:46.000
At this point, we have seen both polynomial long division and synthetic division.
00:26:46.000 --> 00:26:49.700
And so, which is the better method--which one should we use when we have to divide polynomials?
00:26:49.700 --> 00:26:54.200
Now, synthetic division, as you just saw, has the advantage of being fast--it goes pretty quickly.
00:26:54.200 --> 00:26:58.100
But it can only be used when you are dividing by (x - k); remember, it has to be in this form
00:26:58.100 --> 00:27:03.600
of linear things dividing only: x and plus a constant or minus a constant.
00:27:03.600 --> 00:27:08.600
Ultimately, it is just a trick for one very specific kind of problem, where you have some long polynomial,
00:27:08.600 --> 00:27:13.100
and you are dividing by a linear factor--by something x ± constant.
00:27:13.100 --> 00:27:18.600
Long division, on the other hand, while slower, is useful in dividing any polynomial.
00:27:18.600 --> 00:27:20.900
We can use it for dividing any polynomial at all.
00:27:20.900 --> 00:27:25.800
I think it is easier to remember, because it goes just like the long division that we are used to from long, long ago.
00:27:25.800 --> 00:27:30.700
There is a slight change in the way we are doing it, but it is pretty much the exact same format.
00:27:30.700 --> 00:27:35.400
How many times does it fit it? Multiply how many times it fits in by what you started with, and then subtract that.
00:27:35.400 --> 00:27:40.100
And just repeat endlessly until you get to the end.
00:27:40.100 --> 00:27:42.900
And then, lastly, it is connected to some deep ideas in mathematics.
00:27:42.900 --> 00:27:48.200
Now, you probably won't end up seeing those deep ideas in mathematics until you get to some pretty heavy college courses.
00:27:48.200 --> 00:27:56.100
But I think it is really cool how something you are learning at this stage can be connected to some really, really amazing ideas in later parts of mathematics.
00:27:56.100 --> 00:27:58.700
So, personally, I would recommend using long division.
00:27:58.700 --> 00:28:02.300
I think long division is the clear winner for the better one of these to use,
00:28:02.300 --> 00:28:08.700
unless you are doing a lot of the (x - k) type divisions, or the problem specifically says to do it in synthetic.
00:28:08.700 --> 00:28:14.600
If your teacher or the book says you have to do this problem in synthetic, then you have to do it in synthetic, because you are being told to do that.
00:28:14.600 --> 00:28:18.900
But I think long division is easier to remember; it is more useful in more situations;
00:28:18.900 --> 00:28:21.600
and it is connected to some really deep ideas that help you actually understand
00:28:21.600 --> 00:28:24.200
what is going on in mathematics, as opposed to just being a trick.
00:28:24.200 --> 00:28:28.700
Honestly, the only reason we are learning synthetic division in this lesson--in this course--
00:28:28.700 --> 00:28:31.700
is because so many other teachers and books teach it.
00:28:31.700 --> 00:28:34.900
I personally don't think it is that great.
00:28:34.900 --> 00:28:38.000
It is a useful trick; it is really useful in the specific case of linear division.
00:28:38.000 --> 00:28:41.600
If you had to do a lot of division by linear factors, it would be really great.
00:28:41.600 --> 00:28:46.700
But we are just sort of seeing that we can break up polynomials, so I think the better thing is long division.
00:28:46.700 --> 00:28:51.400
It is easier to remember; you can actually pull it out on an exam after you haven't done it for two months,
00:28:51.400 --> 00:28:57.100
and you will remember, "Oh, yes, it is just like long division," which by now is burned into your memory from learning it so long ago.
00:28:57.100 --> 00:29:03.100
And so, synthetic division is really just watered-down long division; I would recommend keeping long division in your memory.
00:29:03.100 --> 00:29:08.200
It is interesting; it is not that hard to remember; it is useful in any situation; and it is connected to some deep stuff.
00:29:08.200 --> 00:29:11.300
And synthetic division is really only useful for this one specific situation.
00:29:11.300 --> 00:29:16.400
So, it is really just a trick; I am not a big fan of tricks, because it is easy to forget them and easy to make mistakes with them.
00:29:16.400 --> 00:29:20.700
But long division is connected to deep ideas, and it is already in your memory; you just have to figure out,
00:29:20.700 --> 00:29:24.000
"How do I apply that same idea to a new format?"
00:29:24.000 --> 00:29:30.800
All right, let's see some examples: Let f(x) = 2x³ + 4x² - 50x - 100.
00:29:30.800 --> 00:29:37.100
Use the fact that f(-3) = 32, and f(-1) = -48, to help you guess a root.
00:29:37.100 --> 00:29:42.000
This sounds a lot like the intermediate value theorem: notice, 32 starts positive; this one is negative.
00:29:42.000 --> 00:29:47.900
So, that means that between these two things, at -3, we are somewhere really positive.
00:29:47.900 --> 00:29:56.000
At -1, we are somewhere really negative; so we know that somewhere on the way, it manages to cross; so we know we have a root there.
00:29:56.000 --> 00:29:59.900
So, how are we going to guess it? Well, we might as well try the first thing that is in the middle of them.
00:29:59.900 --> 00:30:06.800
So, let's give a try to f(-2); now notice, there is no guarantee that f(-2) is going to be the answer.
00:30:06.800 --> 00:30:17.300
It could be f(-2.7); it could be f(-1.005); it could be something that is actually going to require square roots to truly express.
00:30:17.300 --> 00:30:22.900
But we can get a better sense of where it is; and we are students--they are probably going to make it not too hard on us.
00:30:22.900 --> 00:30:25.800
So, let's try -2; let's guess it; let's see what happens.
00:30:25.800 --> 00:30:37.000
We plug in -2; we have 2 being plugged in, so (-2)³ + 4(-2)² - 50(-2) - 100.
00:30:37.000 --> 00:30:42.300
OK, f(-2) is going to be equal to 2 times...what is -2 cubed?
00:30:42.300 --> 00:30:53.000
-2 times -2 is 4, times -2 is -8; keep that negative sign; plus 4 times -2 squared (is positive 4);
00:30:53.000 --> 00:31:01.500
minus 50 times -2; these will cancel out to plus signs; we will get 50 times 2, which is 100; minus 100.
00:31:01.500 --> 00:31:05.600
So, those cancel out (-100 + 100).
00:31:05.600 --> 00:31:12.800
2 times -8 is -16, plus 4 times 4 is 16; they end up being not too difficult on us.
00:31:12.800 --> 00:31:26.300
And sure enough, we get = 0; so we just found a root: f(-2) = 0, so we have a root, or a zero, however you want to say it, at x = -2.
00:31:26.300 --> 00:31:27.500
Great; there is our answer.
00:31:27.500 --> 00:31:33.000
All right, Example 2: f(x) is an even-degree polynomial, and there exists some a and b,
00:31:33.000 --> 00:31:37.400
such that f(a) and f(b) have opposite signs (one positive, the other negative).
00:31:37.400 --> 00:31:42.300
Why is it impossible for f, our polynomial, to have just one root?
00:31:42.300 --> 00:31:45.500
OK, so to do this, we need to figure out how we are going to do it.
00:31:45.500 --> 00:31:52.000
Well, we first think, "Oh, one positive; the other negative; that sounds a lot like the intermediate value theorem that they just introduced to us."
00:31:52.000 --> 00:31:54.000
So, it is likely that we are going to end up using that.
00:31:54.000 --> 00:32:04.200
Let's think in terms of that: f(a) and f(b)...we have two possibilities: f(a) could be positive, while f(b) is negative;
00:32:04.200 --> 00:32:12.500
or it could end up being the case that f(a) is the negative one, while f(b) is the positive one.
00:32:12.500 --> 00:32:15.200
They didn't tell us which one; so we have to think about all of the cases.
00:32:15.200 --> 00:32:17.400
Now, how can we see what this is?
00:32:17.400 --> 00:32:21.300
We have an even-degree polynomial...let's start doing this by drawing.
00:32:21.300 --> 00:32:30.900
We could have a world where we have a positive a (there is some a here, and then some b here,
00:32:30.900 --> 00:32:41.400
where it is negative); and then we could also have another world where we have f(a) start as being negative somewhere,
00:32:41.400 --> 00:32:45.700
and then f(b) is positive somewhere; they don't necessarily have to be on opposite sides of the y-axis.
00:32:45.700 --> 00:32:48.500
But we are just trying to get an idea of what it is going to look like.
00:32:48.500 --> 00:32:52.000
We know that we are somewhere on the left, and then we go up to the right.
00:32:52.000 --> 00:32:53.400
So, how is this going to work?
00:32:53.400 --> 00:32:58.300
And then, we could draw in a polynomial now; we could try to draw in a picture.
00:32:58.300 --> 00:33:00.900
And we might say, "OK, we have a polynomial coming like this."
00:33:00.900 --> 00:33:03.700
And we remember, "Oh, the polynomial could also come from the bottom."
00:33:03.700 --> 00:33:07.000
So, now we have another two possibilities.
00:33:07.000 --> 00:33:11.700
The polynomial is coming from up, or the polynomial is coming from down.
00:33:11.700 --> 00:33:21.600
There is poly<font size="-6">up</font>, if it is coming from above and then going down, or the polynomial that starts below, so poly<font size="-6">down</font>.
00:33:21.600 --> 00:33:24.300
It is coming from the bottom part, and it is going up.
00:33:24.300 --> 00:33:27.200
And maybe that was a little bit confusing as a way to phrase it; but we have one of two possibilities.
00:33:27.200 --> 00:33:35.700
The polynomial is coming in from either the top, or it is coming in from the bottom; polynomial at the top/polynomial at the bottom.
00:33:35.700 --> 00:33:40.200
Those are our two possibilities; OK.
00:33:40.200 --> 00:33:46.600
So, we could be coming from the top; and let's put in our points, as well, again--the same points, positive to negative.
00:33:46.600 --> 00:33:50.500
Or we could be coming from the bottom, like this.
00:33:50.500 --> 00:33:57.400
Then, on our negative to positive, we could have negative down here and positive here.
00:33:57.400 --> 00:34:07.100
And once again, we could be coming from the top or...oops, I put that on the higher one...we could be coming from the bottom.
00:34:07.100 --> 00:34:08.300
So now, let's see how it goes.
00:34:08.300 --> 00:34:14.100
Well, we know for sure that the polynomial has to end up cutting through here, because we are told that it has that value.
00:34:14.100 --> 00:34:17.000
And then, it has to also cut through here.
00:34:17.000 --> 00:34:23.300
In this one, it cuts through here; and then, it has to get somehow to here, so it cuts through here.
00:34:23.300 --> 00:34:28.200
And that is basically the idea of the theorem: we are coming from the bottom, and we go up and come down.
00:34:28.200 --> 00:34:35.400
So, we have already hit more than one root; we have two roots minimum here already.
00:34:35.400 --> 00:34:40.800
What about this one? Well, we go down, and now we have to go up to this one, as well.
00:34:40.800 --> 00:34:44.300
So, we go up, and we are done already; we have two roots for this one.
00:34:44.300 --> 00:34:49.200
In this one, we come up; we go through this one, and then we come up; and we go through this one.
00:34:49.200 --> 00:34:52.700
So, these are the two where we still are unsure what has to happen next.
00:34:52.700 --> 00:34:55.800
Well, remember: what do we know about even-degree polynomials?
00:34:55.800 --> 00:35:04.100
They mentioned specifically that it is an even degree; even degree always means that the two ends,
00:35:04.100 --> 00:35:08.700
if we have it going down this way...then it means that on the right side, it goes down here, as well.
00:35:08.700 --> 00:35:12.600
On the other hand, if it goes up on the left side, then it has to go up on the right side.
00:35:12.600 --> 00:35:18.100
Hard-to-see yellow...we will cover that with a little bit of black, so we can make sure we can see it.
00:35:18.100 --> 00:35:23.200
So, if we are an even degree, they have to be the same direction on both the left and the right side.
00:35:23.200 --> 00:35:28.300
They could both go down, or they could both go up; but it has to, in the end, eventually go off in that way.
00:35:28.300 --> 00:35:31.100
So, who knows what happens for a while here?
00:35:31.100 --> 00:35:36.400
It could do various stuff; but eventually, at some point later on, it has to come back down,
00:35:36.400 --> 00:35:40.600
which means that it has to end up crossing the x-axis a second time.
00:35:40.600 --> 00:35:46.700
So, this one has to be true, as well; the same basic idea is going on over here.
00:35:46.700 --> 00:35:48.500
Who knows what it is going to do for a while.
00:35:48.500 --> 00:35:53.100
But because it is an even-degree polynomial, we know it eventually has to do the same thing.
00:35:53.100 --> 00:35:57.800
So, it is going to have to come back up; so it is going to cross here and here; so it checks out.
00:35:57.800 --> 00:36:04.000
All of our four possible cases, +/- or -/+, combined with coming from the top or coming from the bottom--
00:36:04.000 --> 00:36:08.500
the four possible cases--no matter what, by drawing out these pictures, we see that it is impossible,
00:36:08.500 --> 00:36:12.100
because it is either going to have to hit the two just to make it there;
00:36:12.100 --> 00:36:18.100
or because it has the even degree, it is going to be forced to come back up and reverse what it has done previously.
00:36:18.100 --> 00:36:21.800
And we are getting that from the intermediate value theorem.
00:36:21.800 --> 00:36:31.200
Great; Example 3: Let f(x) = x⁵ - 3x⁴ + x³ - 20x + 60, and d(x) = x - 3.
00:36:31.200 --> 00:36:34.900
And we want to use synthetic division to find f(x)/d(x).
00:36:34.900 --> 00:36:42.400
All right, the first thing to notice: x⁵, x⁴, x³...there is no x²!
00:36:42.400 --> 00:36:46.800
So, we need to figure out what x² is, so we can effectively put it in as 0x².
00:36:46.800 --> 00:36:52.800
Remember: the coefficient that must be on the x² to keep it from appearing, to make it disappear, is a 0.
00:36:52.800 --> 00:36:57.800
So, what we really have is the secret 0x² + 60, because we have to have coefficients
00:36:57.800 --> 00:37:02.800
for every single thing, from the highest degree on down, to use synthetic division.
00:37:02.800 --> 00:37:09.300
So, what is our k? Well, it is x - k for synthetic division; so our k equals 3.
00:37:09.300 --> 00:37:15.400
We have 3 here; and now we just need to place in all of our various coefficients.
00:37:15.400 --> 00:37:21.400
Our various coefficients: we have a 1 at the front; we have a -3 in front of x⁴;
00:37:21.400 --> 00:37:27.300
we have a hidden 1 in front of the x³; we have a 0 on our completely-hidden x².
00:37:27.300 --> 00:37:33.900
We have a -20 on our x, and we have a 60 on the very end for our constant.
00:37:33.900 --> 00:37:37.400
That is all of them; we have made it all the way out to the constant.
00:37:37.400 --> 00:37:40.400
So remember, on vertical arrows, when we go down, we add.
00:37:40.400 --> 00:37:50.800
So, it is adding on vertical arrows: 1 + _ becomes just 1; and then, on these, it is multiplying by whatever our k is.
00:37:50.800 --> 00:37:59.300
So, 1 times 3...we get 3; we add -3 and 3; we get 0; 0 times 3...we get 0 still;
00:37:59.300 --> 00:38:11.100
0 + 1 is 1; 1 times 3 is 3; 0 + 3 is 3; 3 times 3 is 9; -20 + 9 is -11; -11 times 3 is -33; positive 27.
00:38:11.100 --> 00:38:19.300
Now, remember: the very last spot is always the remainder; so this right here is our remainder of 27.
00:38:19.300 --> 00:38:23.900
From there on, -11 is our constant; we work from the right to the left.
00:38:23.900 --> 00:38:30.800
Then come our x coefficient, our x² coefficient, our x³ coefficient, and our x⁴ coefficient.
00:38:30.800 --> 00:38:36.100
And it makes sense that it is going to be one degree lower on the thing that eventually comes out of it, the quotient.
00:38:36.100 --> 00:38:42.800
So, we write this thing out now; we have x⁴ + 0x³, so we will just omit that;
00:38:42.800 --> 00:38:51.000
plus 1x² + 3x - 11; but we can't forget that remainder of 27.
00:38:51.000 --> 00:38:58.000
So, we have a remainder of + 27; but the remainder has to be divided, because that is the one part where it didn't divide out evenly.
00:38:58.000 --> 00:39:08.100
So, 27/(x - 3)--that is what we originally divided by; this is f(x)/d(x).
00:39:08.100 --> 00:39:11.100
Great; and that is our answer, that thing in parentheses right there.
00:39:11.100 --> 00:39:18.100
And now, if we wanted to do a check, we could come by, and we could multiply by x - 3.
00:39:18.100 --> 00:39:22.200
If you divide out the number, and then you multiply it back in, you should be exactly where you started.
00:39:22.200 --> 00:39:31.000
So, we multiply by x - 3; for x⁴, we get x⁵; minus 3x⁴...x²...
00:39:31.000 --> 00:39:44.000
+ x³ - 3x² + 3x(x)...+ 3x²...+ 3x(-3), so - 9x; -11(x), so -11x...
00:39:44.000 --> 00:39:54.000
minus 3 times...-11 times -3 becomes positive 33; and then finally, all of 27/(x - 3) times (x - 3)...
00:39:54.000 --> 00:40:00.200
As opposed to distributing it to the two pieces, we say, "x - 3 and x - 3; they cancel out," and we are just left with 27.
00:40:00.200 --> 00:40:03.700
Now, we work through it, and we check out that this all works.
00:40:03.700 --> 00:40:08.400
So, we have x⁵; there are no other x⁵'s, so we have just x⁵ that comes down.
00:40:08.400 --> 00:40:14.600
3x⁴...do we have any other x⁴'s?...no, no other x⁴'s, so it is - 3x⁴.
00:40:14.600 --> 00:40:20.400
x³: do we have any other x³'s?...no, no other x³'s, so it is + x³.
00:40:20.400 --> 00:40:25.100
- 3x²: are they any others?...yes, they cancel each other out, so it is 0x².
00:40:25.100 --> 00:40:33.400
- 9x - 11x; that becomes - 20x; let's just knock them out, so we can see what we are doing; + 33 + 27 is + 60.
00:40:33.400 --> 00:40:42.600
Great; we end up getting what we originally started with; it checks out, so our answer in red is definitely correct.
00:40:42.600 --> 00:40:48.800
So, remember: that remainder is the one thing where it didn't come out evenly, so it has to be this "divide by,"
00:40:48.800 --> 00:40:53.200
whatever your remainder is here, divided by the thing you are dividing by,
00:40:53.200 --> 00:40:56.300
because it is the one thing that didn't come out evenly.
00:40:56.300 --> 00:41:02.300
All right, the final example: Find all roots of x⁴ - 2x³ - 11x² - 8x - 60
00:41:02.300 --> 00:41:06.200
by using the fact that x² + 4 is one of its factors.
00:41:06.200 --> 00:41:09.600
The first thing that is going to make this easier for us: if we want to keep breaking this down into factors,
00:41:09.600 --> 00:41:16.100
if we want to find the answers, if we want to find what it is--we have to factor it, so that we can get to the roots.
00:41:16.100 --> 00:41:20.900
So, we want to factor this larger thing: we know that we can pull out x² + 4.
00:41:20.900 --> 00:41:25.800
If we are going to pull it out, can we use synthetic division?...no, because it is not in the form x - k.
00:41:25.800 --> 00:41:30.200
We have this x², so we have to use polynomial long division.
00:41:30.200 --> 00:41:42.500
So, x² + 4: we plug in x⁴ - 2x³ - 11x² - 8x - 60.
00:41:42.500 --> 00:41:49.100
Great; x² + 4 goes into x⁴...oh, but notice: do we have an x in here?
00:41:49.100 --> 00:41:54.000
We don't, so it is once again + 0x; so let's rewrite this; it is not just x² + 4.
00:41:54.000 --> 00:42:00.300
We can see this as a three-termed thing, where one has actually disappeared; + 0x + 4.
00:42:00.300 --> 00:42:02.800
Notice that they are the same thing; but it will help us see what we are doing.
00:42:02.800 --> 00:42:05.600
How many times did x² go into x⁴? It goes in x².
00:42:05.600 --> 00:42:14.400
But we don't put it here; we put it here, where it would line up for three different terms: 1 term, 2 terms, 3 terms.
00:42:14.400 --> 00:42:20.100
It lines up on the third term over here, -11x, so it will be x² here.
00:42:20.100 --> 00:42:28.900
x² times (x² + 0x + 4): we get x⁴; this is just blank still; + 4x².
00:42:28.900 --> 00:42:33.700
Now, we subtract by that; we put our subtraction onto both of our pieces, so now we are adding.
00:42:33.700 --> 00:42:40.600
x⁴ - x⁴ becomes 0; -11x² - 4x² becomes -15x².
00:42:40.600 --> 00:42:46.400
We bring down our -2x³; we bring down our -8x; so we have everything.
00:42:46.400 --> 00:42:52.600
-2x³ - 15x² - 8x: once again, we just ask,
00:42:52.600 --> 00:42:55.300
"How many times does the first term go into the first term here?"
00:42:55.300 --> 00:43:02.000
-2x³ divided by x² gets us just -2x².
00:43:02.000 --> 00:43:06.000
Oops, I'm sorry: we are dividing by x², because -2 is just x¹.
00:43:06.000 --> 00:43:14.100
So, -2 times x² gets us -2x³; and then, -2x times 4 becomes -8x.
00:43:14.100 --> 00:43:19.000
We subtract by this; we distribute to start our subtraction, so that becomes positive; that becomes positive.
00:43:19.000 --> 00:43:22.900
Now we are adding: -2x³ + 2x³ becomes 0.
00:43:22.900 --> 00:43:31.800
-8x + 8x becomes nothing; and now we bring down the thing that didn't get touched, the -15x² and the -60.
00:43:31.800 --> 00:43:37.200
And we have -15x² - 60; and hopefully, it will line up perfectly.
00:43:37.200 --> 00:43:41.600
In fact, we know it has to line up perfectly, because we were told explicitly that it is one of the factors.
00:43:41.600 --> 00:43:44.600
So, there should be no remainder; otherwise, something went wrong.
00:43:44.600 --> 00:43:48.000
x² + 0x + 4; how many times does that fit into -15x² - 60?
00:43:48.000 --> 00:43:54.200
Once again, we just look at the first part: -15x² divided by x² becomes just -15.
00:43:54.200 --> 00:44:00.300
-15x²...multiplying it out...4 times -15, minus 60; we now subtract by all that.
00:44:00.300 --> 00:44:02.900
Subtraction distributes and cancels those into plus signs.
00:44:02.900 --> 00:44:10.500
We add 0 and 0; we have a remainder of 0, which is good; that should just be the case, because we were told it was a factor.
00:44:10.500 --> 00:44:14.100
So, we get x² - 2x - 15.
00:44:14.100 --> 00:44:18.300
What is our polynomial--what is another way of stating this polynomial?
00:44:18.300 --> 00:44:26.200
We could also say this as (x² + 4)(x² - 2x - 15).
00:44:26.200 --> 00:44:30.500
Let's keep breaking this up: x² - 2x - 15...how can we factor that?
00:44:30.500 --> 00:44:34.400
x² + 4...can we factor that any more?...no, we can't; it is irreducible.
00:44:34.400 --> 00:44:39.300
If we were to try to set that to 0, we would have to have x², when squared, become a negative number.
00:44:39.300 --> 00:44:45.000
There are no real numbers that do that, so that one is irreducible; we are not going to get any roots out of that; no real roots were there.
00:44:45.000 --> 00:44:47.800
x² - 2x - 09 how can we factor this?
00:44:47.800 --> 00:44:54.700
Just 1 is in front of the x², so that part is easy; it is going to be x and x, and then... what about the next part, -15?
00:44:54.700 --> 00:44:57.300
We could factor that into...one of them is going to have to be negative;
00:44:57.300 --> 00:45:04.400
we factor it into 5 and 3; 5 and 3 have a difference of 2, so let's make it -5 and +3.
00:45:04.400 --> 00:45:10.200
We check that: x² + 3x - 5x...-2x; -5 times 3 is -15; great.
00:45:10.200 --> 00:45:15.000
So, at this point, we set everything to 0; x² + 4 will provide no answers.
00:45:15.000 --> 00:45:20.700
x² + 4 = 0...nothing there; there are no answers there.
00:45:20.700 --> 00:45:31.900
x - 5 = 0; turn this one in--that gets us x = 5; x + 3 = 0 (this gives us all of the roots): x = - 3.
00:45:31.900 --> 00:45:36.900
Our answers, all of the roots for this, are x = -3 and 5.
00:45:36.900 --> 00:45:40.700
And we were able to figure this by being able to break down a much more complicated polynomial
00:45:40.700 --> 00:45:45.800
into something that was manageable, something we can totally factor; and we had to do that through polynomial long division.
00:45:45.800 --> 00:45:48.200
All right, cool--I hope you got a good idea of how this all works.
00:45:48.200 --> 00:45:51.300
Just remember: polynomial long division is probably your best choice.
00:45:51.300 --> 00:45:58.200
Just think of it the same way that you approach just doing normal long division with plain numbers that you did many, many years ago.
00:45:58.200 --> 00:46:01.500
It is basically the same thing, just with a slightly different format.
00:46:01.500 --> 00:46:06.400
How many times does it fit in? Multiply; subtract; repeat; repeat; repeat; get to a remainder.
00:46:06.400 --> 00:46:08.000
All right, we will see you at Educator.com later--goodbye!