Transcription
I've been sent a pretty interesting algebra problem, and I think the question isn't properly formed, or it might be missing some piece of information. But it's an interesting question nonetheless, so I thought I'd work it out, and we can talk about why it's not properly formed. So they give us a function, f(x). They say it's a third-degree polynomial of the form ax³ + bx² + cx + d. And they tell us a couple of the zeros of this polynomial. They say that we have zeros at the point (-1, 0) and the point (2, 0). And they tell us we have a y-intercept, y-intercept, at the point (0, -2). And what they ask us is, what is a + b + c + d? So let's see if we can make some headway on this.
So the first thing is to think about what would a third-degree polynomial look like and what are we even talking about when we say zeros. So let me just draw a little bit of a graph. And I don't know exactly what this third-degree polynomial looks like just yet. Let me draw my axes. Now a third-degree polynomial can have as many as three zeros. And zeros are just the points where the function is equal to 0. So, we have a zero at (-1, 0). So that's (-1, 0), that's right there. We have a zero at (2, 0). (1, 2, 0). So those are the two zeros that they've given us. They also told us the y-intercept at (0, -2). So it intersects the y-axis right there. But this guy can have as many as three zeros. Now some of them might be complex. But complex zeros come in pairs of two, the two complex conjugates. I won't go into too much detail here. So this one must have a third real zero, because if the third one was complex, you'd need another complex zero, and you can't have four zeros for a third-degree polynomial. So that third root, it could be sitting some place out here, maybe over here. Or it could even be a repeat of one of these two roots over here. But let's just assume that there's some third root; we don't know what it looks like. Let's say it looks something like-- let's say that third root is sitting out here some place. And then a potential graph, and I could be completely wrong, I'm just guessing, but just to get a sense of what third-degree polynomials look like, and how can a graph intersect the x-axis three times, a potential graph might look something like this. Curves and bam, hits the other zero, hits the y-intercept, and then goes back up like that. It might go like that, it might go the other way, go like that. Up, and then down like that. Well, there's many ways you could draw something that essentially curves twice to intersect these three points and also that y-intercept, but I'm not going to go into all of those right now. But let's see if we can figure out the coefficients here.
So the key point is me telling you that there must be a third zero here. Let's call that third zero, it's at the point, it's at the point-- let's call it --(r₃, 0). And I'm using the letter r for roots. Roots are the x values of the zeros. So if we say f(-1) = 0, we say that x = -1 is a root. Similarly, we know that f(2) = 0. Or we could say x = 2 is a root; these are roots. So when someone tells you a zero, they're also essentially telling you the roots, and then we know there must be a third root at x = r₃. So x = r₃. Now, if we have a third-degree polynomial where these three x values make it 0, we can rewrite this third-degree polynomial as-- we can rewrite it as-- I'll do it in a slightly different color --f(x) = a(x + 1)(x - 2)(x - r₃). And we want to put an a out front, and I'll tell you in a second why I'm putting this a out front. Because if you were to multiply just the x terms here, when actually do your distributive property and multiply out these binomials, you'll just get x³, but we had an ax³. So you just needed a constant a there to make this an ax³. Now, why did I write it out like this? Well, if you put a -1 in here, what's going to happen? This term is going to be -1 + 1. It's going to be 0. Who cares what these are or what a is, the whole thing's going to be 0. If we put x = 2, this term right here is going to be 0. Who cares what the other terms or the a is, f(2) is going to be 0. Similarly, when x = r₃, this last term is 0, who cares about anything over here. So we know that this guy up here can be rewritten in this form over here. And so if we can solve for r₃, we can just multiply this thing out and try to do a little bit of pattern matching to figure out what these coefficients are going to be.
Now, the other big clue they gave us is a y-intercept. That point right there, the point (0, -2). They're telling us that f(0) = -2. Well, what's f(0)? f(0)-- I'll write it here --f(0)-- If you put 0 in here, this term becomes 0, this term becomes 0, that term becomes 0, you are just left with the d. So f(0) = d. Or we could say that d must be equal to -2. d = -2, so we solved at least one of the coefficients. The constant term. d = -2. Now, if d = -2, that means all of these constant terms, when you were to multiply out this expression, and I would include the a, because the a is scaling everything, that must be equal to -2. Let me be clear here. So let me rewrite this expression. This is equal to, I'll just multiply this a times this last term. We do it in any order you want, so this is the same thing as (x + 1)(x - 2)(a - ar₃). This is the exact same thing as this thing up here, I'll just write this is with f(x) =, and this is the exact same thing as that up there. Now, to solve for r₃, or attempt to solve for r₃, you just have to realize that when you multiply this thing out, the way to get the constant term over here, or the d over here, that's going to be a product of the constant terms in our expressions. Because if I introduced any products with any x terms, you're going to get one of these other terms over here. The ax³ term is generated from that term, that term, and that term, being multiplied together. And then the constant term is multiplied by the constant terms to be multiplied together. And then the two in the middle are multiplied by different combinations of constant terms. And you'll see that if you actually want to multiply this out. But if you take my word for it, that times that times that times, actually let me write it this way. That times that times that, have got to be equal to d; we can then attempt to find some relationship between a and d, which is -2. So we could say 1 * -2 * -ar₃, these two minuses will cancel out. It is going to be equal to our d term. It's going to be equal to -2, and so if we simplify that a little bit we get 2ar₃ = -2, divide both sides by 2a and you get r₃ = -1/a. So we haven't solved for the third root, we've just gotten it in terms of, in terms of our coefficient a. But let's see if we can still use that in some useful way.
So if r₃ = -1/a, this term right here becomes what? Let me rewrite it. Let me rewrite it. All right, I'll do it in blue. f(x) = a(x + 1)(x - 2)(x - r₃). Let me do it over here. -a * r₃ = -a * -1/a. So that's a - * a -. That's a +. The a's cancel out. You just get a 1. So this term now if we substitute r₃ with this will just turn into a this will just turn into a + 1. -a * -1/a = +1. +1. We just substituted that for that to get this right here. Now let's multiply this thing out. Let's actually do the algebraic multiplication and see what we get. So first I'm going to multiply these two terms. I'm going to do it in a different color. Let me multiply those two terms right there. So this is going to be equal to x², x² + 1x - 2x. So that's -x, right? 1 * x is x, -2 * x is -2x. So it's -x, and then -2, just multiplied these two binomials out right here. And then I'm going to have to multiply that times ax + 1, so I'm going to write the ax + 1 down here just to save space. ax + 1. And now we just do a little bit of algebraic multiplication. 1 * -2 is-- I'll do it in magenta ---2. 1 * -x is -x. 1 * x² is x². Now ax * -2 is -2ax. -2ax. ax * -x is -ax², right? -ax² and then ax * x² is ax³. ax³. Then we add everything together, and we get our f(x) = ax³ --let me scroll to the left a little bit --ax³. x² - ax². So that's +1 - a, (1 - a)x². And then we have these two added together so this is -1 - 2a or we could call this equal to-- so -1 - 2a --or we could say -(1 + 2a)x and then we have our -2, which makes sense because that is our d. That's our d. So this is really the solution. This is the form that our equation is going to take, and the reason why I said in the beginning of the problem that the equation or the problem wasn't well defined is that I can pick any real number a and this equation will satisfy these conditions that we were given. So I suspect that I wasn't given all of the constraints on this problem. Maybe there was a third constraint that a = 1, or b = something else, or a + b = something else. But just using these constraints, just using these constraints, the equation will take this form. So, if it was in this form, of course this is a, this is b, which would be 1 - a, that's b, that's a, and then this right here is equal to c. So we could pick, you know, we could pick one particular choice if we just want to get an answer to their problem, but it might not be the answer they were looking for because they might have picked a different a. But if you just pick a simple a = 1, then what's b equal to? Then b = 1 - a, which is equal to 0. Then b = 0, and then c = --actually c would include this minus sign right there --c = -(1 + 2a). So 2a is 2, 1 + 2 is 3. Put a minus sign. So c = -3. And of course, no matter what we do, d is going to be equal to -2. So if we pick a = 1, the answer here to what is a + b + c + d, would be equal to 1 + 0 + (-3) + (-2) = -4. That's another possibility, but we just as easily could have picked, you know a could have been something, it could be a fraction, it could be anything. It could be, you know a could be -1. a could be equal to -1, if a = -1, then b = 2, then b = 2, right? 1 - (-1). And then c = 1 - 2, which is -1, but then you have a minus out front. So then c = 1, and then d is always going to be equal to -2. d = -2. And then you have a + b + c + d would be equal to, well these would cancel out, you'd get them equal to 0. So, we don't know exactly what answer the problem writer was looking for, but it is a pretty interesting problem. And then you know, just out of curiosity, it-- let's take this first example that we took --where a = 1, we could then look at what our root must be. If a = 1, then our third root is that r₃ would be equal-- that's just for this situation right here --then r₃ = -1 / 1 = -1, which means that we would have a repeated root, we would have a repeated root, right there. We wouldn't necessarily have a distinct root out there. And if we're curious about what that would look like, let's look at the situation where a = 1. Actually maybe we will look at both of them. When a = 1, our equation is x³ --where is the x button --x³, there's no x² term, -3x - 2. So let's grab both of them simultaneously. So that's this one, where we picked this choice of coefficients. And then let's do this choice of coefficients, if we have, if we have -x³, right? a = -1 + 2x², and then we have, +x, c is 1, +x - 2. So those are two graphs. Let's see what they look like. Let's graph them. So our first one, there we go, we hit that point, that's our double root. Notice, both of these two graphs meet our two constraints. They both have a root there at -1. They both have a y-intercept at y = -2. And they both have a root at x = +2 right there. Now the second one right here had a separate distinct third root, while the first one, here, has a double root right over there. Anyway, I actually think this problem was somewhat more interesting by the fact we had to actually look at the different solutions to it. And there's actually an infinite number of solutions to this based on what you choose for your a.