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5. For which positive integers d does the polynomial P(X) Xd + Xd-1 X+ have a rational root? Prove your answer:...

Question

5. For which positive integers d does the polynomial P(X) Xd + Xd-1 X+ have a rational root? Prove your answer:

5. For which positive integers d does the polynomial P(X) Xd + Xd-1 X+ have a rational root? Prove your answer:



Answers

Show that the polynomial does not have any rational zeros. $$P(x)=x^{50}-5 x^{25}+x^{2}-1$$

Okay, It's time. They gave us the polynomial X to the 50 finance five X to the 20 last like squared minus one. Okay. And they tell us the show. There are no possible rational euros for this. Well, mother, Impossible. So there are no national zeros for this. And, well, if we've been using, the methods have been using so far. And you know how to perform synthetic division properly. Which means you have to fill in zeros for powers that don't appear there. And the brother's place. Before you perform it, you would immediately screamed. There is no way I am going to be for synthetic division on this quick mathematics and go into Lake. No, I don't know. It felt method of mathematics. But over the polar opposite of mathematics is for you. I mean, it's probably like No, I think so. That method won't work. Synthetic divisions planning upper balance and lower runs. I don't have a panic attack. Aah! We can use the rational zeros there to get the passable, rational zeros will be the device. Is this negative one and one over. The possible devices of the meeting terms coefficient. That's just one. So we divided by that and we see what they will get known like possible solutions. So they're actually two possible rational solution hoping only empty possible. How do we determine it has no rational zero. We'll show that out of the list of possible ones they all fail. This reasoning is kind of like you have three boxes and one of them could contain, like, good bars of gold. And you want to show that none of the boxes contain 100 bars of gold? How did you do it? You would open all the boxes and find out the only of sucks. Okay, so let's do that. Slug in P of negative one, we will get negative 1 1 to 50. Even power. It's just one. What about the next 25th? Negative one. Just changes the same too. What about X Squared? Doesn't change, saying we have minus one. What is its end? Up with seven minus one, which is six. Okay, what about one? Open that and we'll get one of its five plus one minus one. What do we have? What? We have sucks. Okay. None of these are zeroes. It is over. There are no rational zeros to this spot on you.

Sorry, guys. My Internet all could not blow the problem, but I'll explain how I did it. So you first cop down this equation, which is the main equation, and then you get the peace and Q's. So the P remember is this number in the queue is the number that is next to this, which is one basically so then impossible serious are plus minus burek you right off the factors. But the only factor one is one. So the only possible series are plus minus one. If we try plus one rights you put with synthetic division, right? What you're gonna do is you put like every coefficient of the equation off this main equation. And then you bring down the one you want to play one by one. You put it here. You have them native four. What? Supplied by 11 by negative for you. But I hear you got them. You could get the now, since this last term is not Syria, we know that one is not syrup. It would do it for a negative one real quickly. Sorry, this numbers are very bad. I'm just gonna regret it. This are the coefficients of the main equations or 1 to 51 native ones. This one, if you just bring it down the first time you moved above a native one, so and you at them since native. $66.91 or native six times native 16 and you got them a seven night of seven native eight. So you know, this is not serious, that we're proves that it's not that this to are not serious. So in the final answers that there's no serious.

In this problem we have the polynomial B. X. As expert. Five plus X cube plus two X plus one. Now we have the zero as one. Do you want ask one? Bless one plus two plus one. Which is thanks be to us to power five plus two Q. That's too square. That's one comes us 25. So let us find its differential earache X. Which has given us finex powerful. Yes. Three X. Square. That's too I expect for just and X squared both. Her always positive. So this is a positive quantity which means that B. X. S increasing function which means that its graph well look something like this. So it will have a route graph looks something like this. And it will have a route somewhere on the negative side because the zero is positively one is positive and all of the values are positive after zero. It will have no sign change in P. X. After X equals to zero. Which means that it does not have any positive. No positive. Really? Mhm. That's all.

Okay, so you have X cubed plus p x um minus p Q equals zero, where p and Q are prime numbers. So that means that the only possible rational roots need to be factor that these so that could be plus or minus one plus or minus P plus or minus. Q or plus or minus P. Q. Um, but when you try to use synthetic division with any of those, don't forget that you have a zero X squared in there Up. He is a term that, minus p Q um, that you want to get a remainder of zero. Well, if we try and like one like, I'm just gonna go through all the options. And we tried doing it this way. There's no way that you're gonna cancel out. Peak, you okay? Um, I could do the same thing with negative one. That should have been one plus p. Yeah, it's still doesn't kids all but anyway, um and then, ah, this would have a negative p minus one. But you can see how they there's no way to cancel those out. Um, so I just explained why these two don't work. Um, I can quickly explain like P wouldn't work because if you go through it, he squared p squared plus p Um, it's B P Q plus piece. Where'd you see how there's no way to get rid of the Q term that's in here? Um, and I could go. I have to say the same thing about negative p negative p negative p. That's actually the same thing. Um, and then this is negative. P Cube. You see how it ends up really being the same thing? Um, so let me go ahead and race those options, so maybe we can try cute next May. Something nice will happen here. Okay, So cute. I said I would try. Q Q. This is Q squared, but then adding straight down the Q squared plus p and you're left with Q Q plus Peak. You know, that's kind of nice. Now you do cancel out negative P Q plus p. Q. But you still have this Q cubed in here. So that didn't work. You could even try negative que next negative Negative. That's the same. That's the same. But now you get this negative. Que Well, now you just made it worse. So the negative didn't work either. So maybe P. Q. Does work. I'll go ahead, just scroll down, do another one. By the way, these are all the possible rational roots, and so far, six of them I failed, adding straight down. Well, now you get P squared Q squared that when you add straight down and now when you multiply that you're gonna get P cubed Q Q plus P Squared Que Square que. Anyway, nothing cancels there. Do you get a remainder of zero? Absolutely not. And, ah, same thing. If you had negative, she has thrown some negatives. But you really end up with same atrocity where nothing, you cannot get that. So we went through all eight possibilities and none work, so there's no rational roots.


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