Show that $x-1$ is a factor of $P(x)$ where $P(x^5)+xQ(x^5)+x^2R(x^5)=\big(x^4+x^3+x^2+x+1\big)S(x)$ For the problem below, my first step was to rewrite the right side as  $$\big(x^4+x^3+x^2+x+1\big)S(x)=\frac{x^5-1}{x-1}S(x)$$  From there I can isolate $P(x^5)$ which gives me  $$P(x^5)=\dfrac{S(x)(x^5-1)-x(x-1)\big(Q(x^5)-xR(x^5)\big)}{x-1}$$  I'm not sure how to move forward.  To get $P(x)$ should I rewrite all the $x^5$ as $x$ and take the $5^{th}$ root of each $x$?  The questions prior used the idea of roots of unity so perhaps I can use that somehow.    Thanks
If $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$ $P,Q,R,S$ are polynomials such that: $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$   I thought a lot on this but no result!! By the way,one idea is to insert some values for $x$ and try to produce a system of equations for the given polynomials,but I'm not sure it works.
I have a DEM that I would like to smooth or generalize to remove topographic extremes (chop off peaks and fill-in valleys). Ideally, I would also like to have control over the radius or level of "blurriness". In the end, I will need a set of rasters that range from slightly blurry to really blurry. (Theoretically, the blurriest would be a constant raster of the arithmetic mean of all values).  Are there any tools or methods I can use (based on Esri, GDAL, GRASS)?   Do I need to home bake my own  routine?   Could I use a low-pass filter (e.g. ArcGIS's ), and if so, would I need to run it a bunch of times to get the effect of a large radius? Problem: If   $$C_0+\frac{C_1}{2}+\cdots + \frac{C_{n-1}}{n}+\frac{C_n}{n+1} =0,$$  where $C_0,...,C_n$ are real constants, prove that the equation  $$C_0+C_1x+\cdots +C_{n-1}x^{n-1}+C_nx^n=0$$  has at least one real root between $0$ and $1$.  Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 4. Say person A and person B are having a discussion/argument.   Person A tells person B that his views are arrogant, or closed-minded, or some other derogatory adjective. But in fact it is person A's views that are arrogant, closed-minded, etc. Person B's views are usually the exact opposite of what person A is accusing him of.  This sort of situation happens a lot when discussing politics or religion, for example.  What do you call person A, for calling person B something that he himself is?  I guess, for the most part, person A doesn't actually realize that he is the one that the adjective applies to, and not person B. Maybe there is a different name if person A knows the adjective also applies to him. What is the difference between ANSI/ISO C++ and C++/CLI? Remove OTA downloaded (but not installed) iOS 8 I've downloaded the new iOS 8 update on my iPhone 5 but am getting second thoughts about actually installing it.  Is there a way for me to delete that download or am I at the point of no return? Using bm with eulervm gives inconsistent ‘’ characters I prefer to use bm package to bold math. However, I noticed that when used with eulervm, this gives inconsistent results for ‘&lt;’ and ‘>’ characters.  I wonder what causes it and can the usage bm be fixed to get proper ‘&lt;’.  \documentclass{article}  \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc}  %\usepackage{amsmath}  \usepackage[euler-digits,euler-hat-accent]{eulervm}  \usepackage{bm}  \begin{document}      \texttt{\hphantom{mathbold}\llap{bm}}: \quad $ 1 + 1 &lt; 3; \quad \bm{1 + 1 &lt; 3} $      \texttt{mathbold}: \quad $ 1 + 1 &lt; 3; \quad \mathbold{1 + 1 &lt; 3} $  \end{document} Articles: When do I use "a", "the", or "__"? Do I need a UK visa? How and where to apply? I live in Germany since May 2013 as a spouse of German citizen. In Nov 2015 we divorced and I have German residence permit till Jan 2018. We have a son in this marriage and legally I have the right to live in Germany indefinitely. I would like to visit my brother and sister who are British citizens living in UK. What is the visa procedure for me to travel to UK for a week.  Do I need a visa to enter UK ? Label equation with a symbol I would like to label an equation with an arbitrary symbol, rather than a number, or letter.  Is this possible? I want to pose a question such as:  What is your name? Joe  How would I accomplish this using Console.WriteLine to also wait for the response on that same line instead of it being broken into:  What is your name? Joe Why doesn't Quicksilver create sonic booms? How do you detect a network disconnection when downloading a file in java? How to install Gnome 3 without breaking Unity? I tried installing gnome 3 , but it broke Unity and I had to reinstall Natty from scratch.  Does anyone know if there is a way to install Gnome 3 without breaking Unity? Prove universal morphism is unique up to unique isomorphism. On June 28, 2017, Apple  that     macOS High Sierra will be the last macOS release to support 32-bit apps without compromise.   How can I find out which apps on my Mac are 32-bit so that I can see if I need to upgrade them? My script (should) acts differently, depending on the presence of the data in the input stream. So I can invoke it like this:  $ my-script.sh   or:  $ my-script.sh &lt;&lt;-MARK     Data comes...     ...data goes. MARK   or:  $ some-command | my-script.sh   where two last cases should read the data, while first case should notice the data is missing and act accordingly.  The crucial part (excerpt) of the script is:  #!/bin/bash local myData; read -d '' -t 0 myData;  [ -z "${myData}" ] &amp;&amp; {     # Notice the lack of the data. } || {     # Process the data. }   I use read to read input data, then option -d '' to read multiple lines, as this is expected, and the -t 0 to set timeout to zero. Why the timeout? According to  (typing left unchanged; bold is mine):     -t timeout time out and return failure if a complete line of input is               not read withint TIMEOUT seconds.  The value of the TMOUT               variable is the default timeout.  TIMEOUT may be a               fractional number.  If TIMEOUT is 0, read returns success only               if input is available on the specified file descriptor.  The               exit status is greater than 128 if the timeout is exceeded   So I in case 2 and 3 it should read the data immediately, as I understand it. Unfortunately it doesn't. As -t can take fractional values (according to above man page), changing read line to:  read -d '' -t 0.01 myData;   actually reads the data when data is present and skips it (after 10ms timeout) if it is not. But it should also work when TIMEOUT is set to real 0.  Why it actually doesn't? How can this be fixed? And is there, perhaps, alternative solution to the problem of "act differently depending on the presence of the data"?  UPDATE  Thanks to @Isaac I found a misleading discrepancy between quoted on-line version and my local one (normally I do not have locale set to en_US, so help read gave me translation which I couldn't paste here, and looking up for on-line translation was faster than setting new env---but that caused the whole problem).  So for 4.4.12 version of Bash it says:     If TIMEOUT is 0, read returns               immediately, without trying to read any data, returning               success only if input is available on the specified               file descriptor.   This gives a little bit different impression than "If TIMEOUT is 0, read returns success only if input is available on the specified file descriptor"---for me it implied actually an attempt to read the data.  So finally I tested this and it worked perfectly:  read -t 0 &amp;&amp; read -d '' myData;   The meaning: see if there's anything to read and if it succeed, just read it.  So as to base question, the correct answer was provided by Isaac. And as to alternative solution I prefer the above "read &amp;&amp; read" method. Classification of automorphisms of projective space I start with knowing that two numbers are coprime if:  $n*k + m*j = 1$  So, setting $k = a$ and j = $a+1$ I can solve as follows:  $n*a + m*a + a$  Then,  $a(n+m) + m = 1$  Where can I go from here? Let $\alpha$ be an irrational real number and let $a_j$ be a sequence of rational numbers converging to $\alpha$. Suppose that each $a_j$ is a fraction expressed in lowest terms: $a_j = \alpha_j / \beta_j$. Prove that the $\beta_j$ tend to $\infty$  Attempt: AFSOC that $\beta_j$ does not tend to infinity, then it is bounded by some $M$. We can also find an interval $\alpha \in (k, k+1)$. Let $\epsilon &gt; 0$, $\exists N_0$ such that for all $N &gt; N_0$ we have $|a_N - \alpha|  &lt; \epsilon$. So $|\frac{p_N}{q_N} - \alpha| &lt; \epsilon$. We try to use the fact that $q$ is bounded but fail to derive a contradiction. I do not know how to use the fact that $\alpha$ is between two integers, although it might not be relevant at all.  Ideas? .NET OpenSource projects and strong named assemblies?