Timeline for Compute polynomial $p(x)$ if $x^5=1,\, x\neq 1$ [reducing mod $\textit{simpler}$ multiples]
Current License: CC BY-SA 4.0
21 events
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| Sep 15, 2019 at 9:06 | history | edited | nonuser | CC BY-SA 4.0 |
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| Jun 18, 2019 at 17:43 | comment | added | user193810 | @Aqua: it is a well-written answer, so it deserves the upvotes. It does not answer the question, but apparently Joeblack asked the wrong question, and you gave the answer to the question that should have been asked. :P;) | |
| Jun 18, 2019 at 14:31 | comment | added | nonuser | @Pakk Now 43+ :) and 3- :( | |
| May 27, 2019 at 0:29 | audit | First posts | |||
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| May 18, 2019 at 9:20 | comment | added | nonuser | If $a$ is a double root for $Q$ you can not deduce that from my proof for $P$. So it need not $Q$ to divide $P$. Say $Q(x) = (x-1)^2(x+2)$ and $P(x)= (x-1)(x+2)$ then every zero of $Q$ is also zero for $P$, but $Q$ does not divide $P$.@rb612 | |
| May 18, 2019 at 9:16 | comment | added | rb612 | @MariaMazur what would be the problem? | |
| May 18, 2019 at 8:54 | comment | added | nonuser | No, in that case my proof would not work. @rb612 | |
| May 18, 2019 at 8:48 | comment | added | rb612 | @MariaMazur ah yes, thank you for clearing that up. What I meant more was that wouldn't the claim still work if $a, b, c, d$ were all the same zeros? Then it would have multiplicity 4 and thus it would be divisible by $(x-a)^4$? In other words, is there anything special about having the roots be distinct such that this divisibility holds? | |
| May 18, 2019 at 8:13 | comment | added | nonuser | If $a$ is a zero of $P(x)$, then $P(x)= (x-a)P_1(x)$, so if $a,b,c,d$ are all different zeroes, then we have $P(x) = (x-a)(x-b)(x-c)(x-d)R(x)$, for some polynomial $R$. But $ (x-a)(x-b)(x-c)(x-d) = Q(x)$ and thus a conclusion. @rb612 | |
| May 18, 2019 at 5:25 | comment | added | rb612 | "since all 4 zeroes of $Q(x)$ are different..." what's the necessity of having all the roots be distinct in order to conclude the final statement? | |
| May 16, 2019 at 11:14 | comment | added | user193810 | @smci: Indeed. Hence the OP's questions: "How is the student supposed to think of the suggested method? Is it obvious? How else could one approach the problem?" In my opinion, Maria does not address any of these three questions. She does give an excellent answer to the question how to calculate the remainder of $P$ divided by $Q$. If that question would have been asked, this should have been the accepted answer. | |
| May 16, 2019 at 9:44 | audit | First posts | |||
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| May 16, 2019 at 1:56 | comment | added | smci | @Pakk: actually really you only need to spot the geometric-series identity for $Q(x)$, the rest is fairly obvious if you know complex numbers. | |
| May 16, 2019 at 1:53 | comment | added | smci | @Pakk: it hinges on spotting the standard identity: $Q(x) = 0$ gives the fifth roots of unity (other than x=1). It takes tons more work if you don't spot that; you could solve the quartic $Q(x) = 0$ then notice "Hey these are the other four fifth roots of unity, thus x^5 \equiv 1" then apply the information $x^5 \equiv 1$ to hugely simplify $P(x)$, and thus conclude $Q(x) \mid P(x)$ and remainder = 0. But that would be lots more work. (Yeah the first time I saw this trick back in HS I voiced the same objection about non-obviousness... this is why this identity is so pivotal to factorization) | |
| May 14, 2019 at 13:25 | comment | added | user193810 | @MariaMazur: A bit, but not really. It still feels like you stopped reading the question after "what is the remainder of P(x) divided by Q(x)?". You gave a correct answer to that question, but not to the real question. It is easy to prove that the 'trick' (multiplication by $x-1$) works and is correct, but you ignored the real question, and after your edit it is hidden as a small remark. But whatever, your answer is trivially better than mine, because I did not gave any, so I'll leave it here. | |
| May 14, 2019 at 12:34 | comment | added | nonuser | @Pakk Do you see it now? | |
| May 14, 2019 at 12:33 | history | edited | nonuser | CC BY-SA 4.0 |
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| May 14, 2019 at 11:49 | comment | added | user193810 | The OP understands that after multiplication with $x-1$, the question became simpler, he/she asked how a student should find this trick. This answer is basically just a reformulation of the answer given in the question. I don't see the usefulness of this answer at all... (but it has 14 upvotes, so I might be missing something...) | |
| May 14, 2019 at 3:54 | comment | added | trancelocation | From my perspective this is the best answer. (+1) | |
| May 14, 2019 at 0:23 | comment | added | Dancrumb | I think this is the approach that the examiners would expect from high school students. | |
| May 13, 2019 at 19:07 | history | answered | nonuser | CC BY-SA 4.0 |