College Algebra: Find Polynomial from Zeroes, etc

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So, what we've seen is, with cubic polynomials there will be three roots. There'll be three 0's. Two of them may be imaginary, but there will always be three. So, that sort of raises an interesting question. Suppose that I just tell you the 0's. Would that allow you to figure out which cubic polynomial I'm thinking about? Let's just think about that for a second, visually. Suppose I say to you, "Hey, I'm thinking of a cubic polynomial and it has a 0 right here, right here and right here." Do you know that cubic polynomial for certain?
Well, the answer is no, because there are many cubic polynomials that actually have these as roots. Let me show you some visually. I'm going to show you the graphs of those things. One might go up, down and like that. There's a cubic polynomial that has these as the roots. Let me show you another one. You can have even a steeper one. Look at that one. See, that's a completely different one and yet still has these as a root. You could also imagine one that has like a negative, sort of an unhappy one. Comes down like this, goes up, comes down. So, there are a lot, in fact, infinitely many possible cubics if I just tell you the roots.
However, if I tell you the roots and just one other point, that the function actually satisfies--so, maybe I say, "Okay, it has these three roots and you know what, it also passes through this point right here." So, at 10, I know the function is going to be 7. Then, those four points, believe it or not, completely determine that cubic. There's only one cubic that's going to pass through all those things. Maybe it would look like this. There's only one answer.
So, just knowing the roots, just knowing the 0's and one extra point, you can pinpoint that function explicitly for cubics. So, let me illustrate that with an example. Let's find the cubic polynomial, a third degree polynomial that has it's 0's at x = -3, x = 1, and x = 4. So, it has three 0's, ones at -3, 1 and 4. I have to give you one more piece of information. I'll tell you that if you plug in 2, into the cubic, the thing you get is going to be 30. Now, I'm going to show you how you can find out, for sure, pinpoint, the equation for that particular cubic. The secret is to use the factor theorem again.
So, the factor theorem tells us that x = -3 is going to be a 0, precisely if x - (-3) is a factor. So, what that means is that I know that x - (-3) must be a factor of this. I could do the same thing here. If x = 1 is 0, that means that x - 1 must be a factor. If x = 4 is 0, then x - 4 must be a factor. So, that must be the factorization of f. Except, there may be some big coefficient in front, some number. I'll call it "a". So, I just almost know the polynomial completely. The only thing I have to do now is find "a". How could I find "a"? Well, one way to find a, is just to use this fact. That particular fact is going to pinpoint "a" completely. Watch.
If I let x be 2, so wherever I see an x, I'm going to plug in 2. Not here, that's not an x. That's something else. Wherever I see an x, I'm going to plug in 2, then I know this has to equal 30. That will allow me to solve for "a". Let's try it. f(2), on the one hand it equals "a" times, now this is a 2 and then this is a minus a minus, is a plus, so we have 2 + 3, which is 5, times, put a 2 in here. I see 2 - 1, which is 1. Put a 2 in here, I see 2 - 4, which is about -2. Now, that we know, we're told equals 30. So, if I set that equal to what it equals, 30, I can now solve this for "a". What do I see? I see that, well, this is just -10.
So, I see that -10a = 30. If I divide both sides through by -10, I see that "a" must be -3. So, there is a coefficient, the coefficient is -3. This polynomial is equal to -3(x + 3)(x - 1)(x - 4). That is exactly the polynomial. Now, if you want to know what that is in unfactored form, you can just multiply everything through. First of all, multiply that out. That term right there, for example, would give you an x² + 2x -3. Do you see it? 3x - x is plus 2x - 3, that's that piece. Then you've got to take that whole thing and multiply it through by x - 4. So, you have to distribute that x through each of the three terms. You have to distribute the -4 through each of those three terms. If you do that, I'll just tell you the answer. The answer is going to be something like, something like, -3(x³ - 2x² - 11x + 12).
Now, I'm going to leave it to you, right now as a little challenge, to just check that by multiplying these three things out yourself, and verifying that you get that. Multiply the whole thing through by -3 and there's your answer. Let's try one last one. Suppose I tell you I'm thinking of a cubic polynomial and the roots, well the 0's, are at x = -1, 2, and 4. I need to know one more piece of information. I'll tell you that if you plug in 1, you get 3. Now the question is, "What is the equation for that cubic?"
Let's use the factor theorem. There may be some coefficient out in front, but if -1 is a 0, that means that x - (-1) must be a factor. If 2 is a 0, that means x - 2 is a factor. If 4 is a 0, that means x - 4 is a factor. So, all this is known, everything is known accept for that number in front that may be there. So, what is this? Well, that's there. I know this fact. So, f(1), if I let x = 1, this whole thing has to equal 3. Let's see what we get. Now that's not an x, so I just keep that there, that's "a". Here I have a 1 - (-1). That's 1 + 1, which is around 2. Then I see a 1 - 2, which is -1. Then I see a 1 - 4, which is a -3. That whole thing is combined to gives me 3. That's what we're told, f(1) equals 3. This is f(1).
So, I can solve for "a", because this tells me that I have 6a = 3 and so that implies that a = . So, if a = , I just put a in there and I see what the polynomial is. It equals (x + 1)(x - 2)(x - 4). There's the answer and that's the way I like to think of it. In fact, that is, I think, the best way to look at the polynomial. Okay, maybe some people want you to actually work that out all the way. But they actually want you to work that out and actually figure out what that is, just in terms of the cubic, multiply that out. Let me just do that for you, now. The last one I'll let you do. I'll do this one just to show you that I'm fair about these things.
We have the way out in front. First, I'm going to multiply this out. That's going to give me an x². Then inside I have an x, outside I have a -2x. That's a net gain of -x and then I have a -2. That's all multiplied by x - 4. So, now I've got to multiply everything through by x - 4. Remember how that goes. You have to distribute really carefully. This has to hit everybody. Then that -4 has to hit everybody. So, if you're just careful with it, there will be no problem. X times this, I have a out in front. Okay, x times x² is x³, x time -2 is -2x. I just took the x and just multiplied it through by everything.
Now, I'm going to take the -4 and hit it with everything. So, what do I get? I get -4x². I see now, a plus, a negative and a negative is a plus, 4x and then here I see a plus 8. So, what does that equal? Well, I can combine this a little bit. I have , I've got an x³ and I have some x² here. I have an x², -x² and then a -4x², that gives me a net total of -5 of them, -5x². Then I have a -2x, but then I have a +4x, so that's just +2x. Then I just have a +8 at the very end. So, this is another way to say the answer, or you can distribute now through by the and you'd see the following. f(x) equals x³ - ^5/[2]x² + x +4. You see where I distributed the through? That is the one cubic polynomial, the one cubic polynomial that has the feature, the roots are -1, 2, and 4, and that passes through (1,3).
Polynomial and Rational Functions
Zeros of Polynomials
Finding Polynomials Given Zeros, Degree, and One Point Page [1 of 2]