I'm using the polynomial division method and i got a remainder of 32 but how do I explain in words or mathematically how this justifies my solution.  The previous question has answers based on Properties of Divisibility but I want to know how I can use long polynomial division in my method.  Thanks
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My attempt was suppose $\gcd (n+5, n^2+9) = d$ where $d$ is an integer, hence $d|n+5$ and $d|n^2+9$ but I don't know how could I rearrange this to make it $1$
71274
I am really having trouble working a problem in my course. I will post it and then add what I have thought etc.  Let p be an odd prime, then using the Euclidian Algorithm, calculated the GCD of of $x^{p-1}-1$ and $x^2+1$ in $\mathbb{Z} / p\mathbb{Z}[x]$  Also, how does $p=1 (mod4)$ tell us that  $-1$ is a square in $\mathbb{Z} / p\mathbb{Z}$, infact that $-1$ is a square here if and only if $p=1(mod4)$.  My thoughts:   I know that $$x^{p}-x=(x)(x-1)(x-2)…(x-(p-1))$$ and so that in this sense our roots will be the congruence classes of $1,2,…p-1$  and also that $$x^{p-1}-1=(x-1)(x-2)…(x-(p-1))$$  But I get lost when trying to apply the EA to these polynomials,  like how can I do it when p is arbitrary?  Preferably id like to be able to understand/see how to do the first part first i.e. using division with residue and then move on to the second part. Any insight/hint or solution to this would be great,  Thank you all  Update:  When I try to do division with residue, I just get  $$x^{p-1}-1=(x^2+1)(x^{p-3}-x^{p-5}+x^{p-7}-x^{p-9}...)$$  for example,  $p=7$  $$(x^{6}-1)=(x^2+1)(x^4-x^2+1)$$  for $p=11$  $$(x^{10}-1)=(x^2+1)(x^{8}-x^{6}+x^{4}-x^{2}+1)$$  which for specific cases tells me that 1 is the gcd, but I dont know how to generalize that for the case with arbitrary prime p.
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This question arose while I was tutoring a student on the topic of the Remainder Theorem. Now, the Remainder Theorem tells us that when a polynomial $p(x)$ is divided by a linear factor $(x-a)$, the remainder is simply $p(a)$. However, in this case we have a product of linear factors.   Using the Remainder Theorem we can see that neither $(x-1)$ nor $(x+2)$ is a factor of $f(x)$. Also, if we try to find the remainder using long division, we get a relatively ugly remainder of  $$ 3(14x - 13) $$ I assume this is not the correct approach as all other questions in this topic used the Remainder Theorem. So perhaps there is a more elegant approach?
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The following question was asked on a high school test, where the students were given a few minutes per question, at most:     Given that,   $$P(x)=x^{104}+x^{93}+x^{82}+x^{71}+1$$   and,   $$Q(x)=x^4+x^3+x^2+x+1$$   what is the remainder of $P(x)$ divided by $Q(x)$?     The given answer was:     Let $Q(x)=0$. Multiplying both sides by $x-1$:   $$(x-1)(x^4+x^3+x^2+x+1)=0 \implies x^5 - 1=0 \implies x^5 = 1$$   Substituting $x^5=1$ in $P(x)$ gives $x^4+x^3+x^2+x+1$. Thus,   $$P(x)\equiv\mathbf0\pmod{Q(x)}$$     Obviously, a student is required to come up with a “trick” rather than doing brute force polynomial division. How is the student supposed to think of the suggested method? Is it obvious? How else could one approach the problem?
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I'm working on a math problem but I am having a hard time figuring out the method used by my textbook to make this factorization:  $$x^4 + 10x^3 + 39x^2 + 70x + 50 = (x^2 + 4x + 5)(x^2 + 6x + 10)$$  I've tried to see if this equation can be factored by grouping or by long division to no avail. Any help would be greatly appreciated.
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Let f(x) be a  polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$  $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ . And then there are some question related to this. In the solution it is given that according to the given information, $f(x)$ is a polynomial of degree 4 with leading coefficient 3. $f(x)=3(x-2)(x-3)(x+1)(x+6)+(x^2+1)$ Now what i think is that he got the coefficient 3 using the limits and the rest of the terms$[(x-2)(x-3)(x+1)(x+6)]$ from the respective $f(x)$ values from what was given and also the $(x^2+1)$ was added for the remainders, as they satisfy all the values that have been given to us in the question. I just wanted to know if my approach is right? or is there another elegant way to do it?.
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Why is it that $\forall n \in N$, $2^n$ is not divisible by $3$?  I can prove it easily by induction, but I don't understand the intuition of why this is true. Could anyone please supply the intuition? [This has been answered.]  Moreover, I understand that there's also a proof of this fact that uses modular arithmetic. How does this proof go? [Also answered - it follows immediately from the definition of the modulus operation.]
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I don't understand how to use Fermat's little theorem to find remainders e.g if we are asked to find remainder of $50^{50}$ on division by $13$, what is a and what is $p$ in the formula?  Also I wanted to check can we use both congruence classes as well as Fermat's little theorem to find such remainders? How would be write $50^{50}$ on division by $13$ in terms of congruence classes to find the remainder?  I don't exactly understand what a congruence class is, what does it tell us and how is it related to this?
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This is from Pinter, A Book of Abstract Algebra, p.265.   Given $p(x) \in F[x]$ where $F$ is a field, I would like to show that $p(x)$ divided by $x-c$ has remainder $p(c)$.  This is easy if $c$ is a root of $p$, but I don't see how to prove it if $c$ is not a root.
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What will be the remainder when $f(x^{12})$ is divided by $f(x)$ where : $$f(x) = x^5 + x^4 + x^3 + x^2 + x +1$$  I have already tried but found no idea how I can do this question.
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Fix a nonnegative integer $n$. Let $P(x)$ be a polynomial of degree $n$ (over the real numbers) such that for all $k\in\left\{0,1,\ldots,n\right\}$, we have $P(k)=\dfrac{k}{k+1}$. Find $P(n+1)$.   My attempt:  I consider a new polynomial $Q(x)=(x+1)P(x)-x$. This $Q$ is a polynomial of degree $\leq n+1$ satisfying: $Q(k)=0\quad\forall k\in\{0,1,2,\ldots,n\}$. Therefore $Q(x)=\lambda x(x-1)\cdots(x-n)$ for a certain real $\lambda$. Hence, $Q(-1)=\lambda \cdot (-1)^{n+1}(n+1)!$, but also $Q(-1)=(-1+1)P(-1)+1=1$. Comparing these, we get $\lambda \cdot (-1)^{n+1} (n+1)! = 1$, thus $\lambda=\dfrac{(-1)^{n+1}}{(n+1)!}$. Since $Q(n+1)=\lambda(n+1)!$ then $Q(n+1)=(-1)^{n+1} $, and from $P(x) = \dfrac{Q(x)+x}{x+1}$ we deduce: $$P(n+1)=\dfrac{(-1)^{n+1}+n+1}{n+2}.$$ I doubted that the answer were $P(n+1)=\frac{n+1}{n+2}$ but the term $(-1)^{n+1}$ wasn’t expected. Is there any mistake here? Does someone have an alternative proof?
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I'm supposed to prove that, for every integer $n &gt; 0,$ it is true that $(1 + 2 + ... + n)$ divides $3(1^2 + 2^2 + ... + n^2)$.  Should I use induction?    This was given as an exercise in a chapter about divisibility in my algebra textbook.
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What is the fastest known way for testing divisibility by 7? Of course I can write the decimal expansion of a number and calculate it modulo 7, but that doesn't give a nice pattern to memorize because 3 is a primitive root. I'm looking for alternative ways that can help you decide when a number is divisible by 7 by hand.  I'm sorry if this is a duplicate question, but I didn't find anything similar on the site.
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my question is      If $-27$ is divided by $5$, what would be the remainder?
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Disclaimer: I'm an engineer, not a mathematician  Somebody claimed that $\gcd$ only is applicable for integers, but it seems I'm perfectly able to apply it to rationals also:    $$ \gcd\left(\frac{13}{6}, \frac{3}{4} \right) = \frac{1}{12} $$  I can do this for a number of cases on sight, but I need a method. I tried Euclid's algorithm, but I'm not sure about the end condition: a remainder of 0 doesn't seem to work here.    So I tried the following:    $$\gcd\left(\frac{a}{b}, \frac{c}{d} \right) = \frac{\gcd(a\cdot d, c \cdot b)}{b \cdot d}$$  This seems to work, but I'm just following my intuition, and I would like to know if this is a valid equation, maybe the correct method. (It is in any case consistent with $\gcd$ over the natural numbers.)  I'm not a mathematician, so please type slowly :-)
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Is there a division algorithm for polynomials in R[x], where R is a commutative ring with unity?    All the algebra books I read mention division algorithm for polynomials in F[x], where F is a field. Since the leading coefficient of a non zero polynomial in R[x] is not necessarily invertible, I find it difficult to proceed the same way we do for polynomials in F[x].  Can I get some help?
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Excuse me for the confusing title. I was asked to find $gcd(x^2+3x+2,x^2+x)$  What i did is i factorized both polynomials $x^2+x=(x+1)x$  $x^2+3x+2=(x+1)(x+2)$  So i expected the gcd to be $x+1$   But using the euclidean algorithm i found out the gcd to be $2x+2$. Why is factorizing wrong? Is it because $K[X]$ is not factorial ? Would the euclidean algorithm also work if the polynomials are in $\Bbb Z[X]$ ???