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7

The roots of this polynomial are $$e^{2k\pi i/5}$$ for $k\in\{1,2,3,4\}$. The point is that the roots for $k=1$ and $k=4$ are conjugate complex numbers, as well as the roots for $k=2$ and $k=3$. Furthermore, $$e^{2\pi i/5}+e^{8\pi i/5}=2\cos\left(\frac{2\pi}5\right)$$ and $$e^{4\pi i/5}+e^{6\pi i/5}=2\cos\left(\frac{4\pi}5\right)$$ Then ...

6

$$(x+1)^n=((x-1)+2)^n=2^n+n 2^{n-1}(x-1)+\binom{n}{2}2^{n-2}(x-1)^2+(x-1)^3 \cdot F(x)$$ So the remainder is $$2^n+n 2^{n-1}(x-1)+\binom{n}{2}2^{n-2}(x-1)^2.$$

6

A monic polynomial is any polynomial $f(x)=a_n x^n + a_{n-1} + \cdots + a_1 x^1 + a_0 x^0$ such that $a_n=1$. Therefore, a monic polynomial of degree zero is of the form $f(x) = a_0$ where $a_n = a_0 = 1$ as $n=0$ so they may only take the form $f(x) = 1$.

5

Let $r=s/t$ where $s\in\mathbb{Z}$ and $r\in\mathbb{N}$. Assume $e^r=p/q$ where $p,q\in\mathbb{N}$. Then $$pqt^nJ_n(r)= p^2 t^n A_n\left(\frac{s}{t}\right)+ q^2 t^n B_n\left(\frac{s}{t}\right)\in\mathbb{Z} \tag{1}$$ Note that $$0<J_n(x)\leq\frac{x^{2n}}{n!}\int_{-x}^xe^tdt=\frac{2 x^{2n}\sinh x}{n!}$$ So $$0<pqt^nJ_n(r)\leq \frac{2 pq t^n ... 5 Write 2014 = 2N, where N = 1007 is odd. I claim that$$f(x) = N x^2 + 2 x = 1007 x^2 + 2x$$has the required property. Note that f(x) is odd if x is odd and even if x is even. Hence, if f(i) and f(j) have the same remainder when divided by 2014 = 2N, then i and j have the same parity, and so i-j is divisible by 2. Suppose that ... 4 This is actually a very sloppy simplification of an important and complicated proof of the fundamental theorem of algebra, which secretly proves nontriviality of \pi_1(S^1) behind the curtains (which involves some nontrivial bit of algebraic topology). I'll try to give a more explicit description of what's really going on in the proof by answering your ... 4 This might not be true. See for instance f(X)=X^3-2 (irreducible by Eisenstein) then :$$\dim_{\mathbb{Q}}(\mathbb{Q}[x]/(f(x))=\deg(f)=3 $$However the roots of f are \sqrt[3]{2}, \sqrt[3]{2}j and \sqrt[3]{2}j^2. It is easy to see that F:=\mathbb{Q}[{\sqrt[3]{2}},{\sqrt[3]{2}}j,{\sqrt[3]{2}}j^2] cannot be of dimension 3 over \mathbb{Q} ... 3 I don't think the following could be caracterized as "nasty determinant calculation". I don't know how one can prove the equality without indulging in some computation. Let r=\operatorname{rank}(A) From a well-known theorem, derive that there exists P,Q invertible m\times m and n \times n matrices such that$$A=P\begin{bmatrix}I_r& 0\\ 0 ...

3

Well ordered $\implies$ every chain has a minimum, but $$1>m \implies 1>m>m^2>\dots$$

3

Hint : The desired polynomial $p(x)$ must have $3$ as a root, that means $p(3)=0$. It is not difficult to see that $1$ must be added using this fact.

3

Let's decompose $P$ into distinct monomials, say, $P=a_1+ ... + a_k + b_1 + ... + b_l$ where $a_1+ ... + a_k$ is the lacunary part of $P$. Let $\Phi_\sigma$ be the automorphism on $R[X_1,...,X_n]$ corresponding to a permutation $\sigma\in S_n$. Then, $\Phi_\sigma( a_1 ) + ... + \Phi_\sigma( a_k ) + \Phi_\sigma( b_1 ) + ... \Phi_\sigma( b_l ) = ... 3 You can write the equality (E): $$(x+1)^n=P(x)(x-1)^3+a(x-1)^2+b(x-1)+c$$ where$P(x)$is the quotient and$a(x-1)^2+b(x-1)+c$the reminder of the euclidean division of$(x+1)^n$by$(x-1)^3$. And your problem is to find$a,b,c$. (E) can be seen as an equality between functions. So can you differentiate both side to get further equalities. Make$x=1$in ... 3 Let$A$and$B$be domains with$A\subseteq B$. We say that the ring extension$B/A$is good if the following implication holds: $$\text{For every}\ f\in A\setminus\{0\}\ \text{and}\ h\in B,\ \text{if}\ fh\in A\ \text{then}\ h\in A\,.$$ We claim that if$B/A$is good then$B[x]/A[x]$is good as well. In fact, let$f\in A[x]\setminus\{0\}$and$h\in ...

3

As you said, if $a \neq 0$ we have $f(a^{-1})=0$ which means that $f$ has a root and thus is reducible [$x-a^{-1}$ is a factor]. If $a=0$ you need to factor $x^6+5$. To do this note that $5 \equiv -100 \pmod{7}$.

2

Note that $f(a^{-1})=0$ as you noted before when $a$ is not $0$. Then we know that $(x-a^{-1})$ is a factor. We can do a "complete the sextic" approach on the function as follows: $f(x) = (x-a^{-1})(x^5) = x^6-a^{-1}x^5$ $f(x) = (x-a^{-1})(x^5+a^{-1}x^4) = x^6 + a^{-2}x^4$ Verify that $f(x) = (x-a^{-1})(x^5+a^{-1}x^4+a^{-2}x^3+a^{-3}x^2 + a^{-4}x + 2a) = ... 2 (1) Consider$f(x)=x^n-2$then we'll show that it have no rational root. suppose$x=\frac{p}{q} ~with~gcd(p,q)=1$be a root of$f(x)$.$\implies (\frac{p}{q})^n-2=0\implies p^n-2q^n=0\implies p^n=2q^n\implies q\mid p^n$but$gcd(p,q)=1$and therefore it shows that$q=1$So we have$p^n=2~for~ all~ n\geq2$. Which is a absurd since 2 is prime. ... 2 Let$f\mid g$in$L[x]$. Polynomial long division yields a polynomial$q\in L[x]$with$f = gq$. A closer look at the polynomial long division algorithm shows that the coefficients of$q$are computed by repeatedly applying field operations to the coefficients of$f$and$g$. So if the coefficients of$f$and$g$are in$K$, then the computed coefficients of ... 2 When you are doing the simplification:$\frac{2(x-2)}{(x-2)}$= 2, you are making the claim that$\frac{(x-2)}{(x-2)}$=1. This is true for almost any value of x, for example: x= 5$\frac{(5-2)}{(5-2)}$=$\frac{3}{3}$= 1 However, in the case$\frac{(2-2)}{(2-2)}$=$\frac{0}{0}$, which is undefined. So, that step you do is value for all values of x ... 2 We look only at$P(x)=x^{10}-x^7+x^4-x^2+1$. It is clear at a glance that$P(x)\gt 0$if$|x|\ge 1$. Grouping as$(x^{10}-x^7)+(x^4-x^2)+1$does it. So we look at$|x|\lt 1$. Negative$x$in this range are easy to deal with, so we concentrate on$0\lt x\lt 1$. Since$x^4-x^7\gt 0$, we have$P(x)\gt 1-x^2\gt 0$. 2 Just another way for$(a)$is using the AM-GMs: $$\frac12x^{10}+\frac12x^4 \ge x^7, \quad \frac12x^4+\frac12 \ge x^2$$ $$\implies x^{10}-x^7+x^4-x^2+1 \ge \frac12+\frac12x^{10}>0$$ and similarly for$(b): $$\frac12x^4+\frac12 \ge x^2, \quad \frac12x^4+\frac32+\frac32+\frac32 \ge 2\times 3^{3/4}x> 3x$$ Not always applicable, of course. Also you ... 2 Hint: $$f(x)=(x-3)(x+1)(x-a)$$ $$f(4)=5(4-a)=30$$ 2 You don't need to solve the third degree equation.\begin{align}(2-x)^2\color{red}{(-2-x)}-\color{red}{(-2-x)}&=\color{red}{(-2-x)}((2-x)^2-1)\\&=-(2+x)((x-2)^2-1^2)\\&=-(x+2)(x-2-1)(x-2+1)\\&=-(x+2)(x-3)(x-1)\end{align} 2 Hint: $$a^2-b^2=(a-b)(a+b)$$ Apply this to $$(2-x)^2 \cdot (-2-x) - (-2-x) = - (x+2)((2-x)^2-1^2)$$ 2 There is a trick. Note that(-2-x) = -(x+2)is a factor of both summands so \begin{align*} (2-x)^2 \cdot (-2-x) - (-2-x) & = -(x+2)\cdot((2-x)^2 - 1) \\ & = -(x+2)(x^2 - 4x + 4 - 1) \\ & = -(x+2)(x^2-4x+3) \\ & = -(x+2)(x-3)(x-1) \end{align*} Where we only had to factor the quadraticx^2-4x+3$2 Let the original polynomial be$p(x)$. By Remainder Theorem, the remainder when dividing$p(x)$by$(x-a)$is$p(a)$. Applying that here, the remainder when dividing$p(x)$by$(x-3)$is$p(3) = -1$. To get a remainder of$0$, we simply add one to the polynomial ($-1 + 1 = 0$). It's that simple. 2 Any function of the form $$y = ( a_1x-b_1)^2(a_2 x -b_2)^2 ( a_3 x - b_3)^2$$ will have your triple well form with centres$b_i/a_i$, this fixes 3 free variables. To determine the heights, you want the derivative at the peaks to be zero. Solve the resulting equations and you'll have your conditions. In your example we have $$y = x^2 (x-3)^2(x+3)^2$$ You ... 2 Let me suppose that you want $$A(x)=\sum_{n=0}^\infty a_i x^i$$ So, let us mutliply both sides by the denominator $$8+14x-50x^2=(1-7x^2+6x^3)\sum_{n=0}^\infty a_i x^i$$ Now, decompose the product $$8+14x-50x^2=\sum_{n=0}^\infty a_i x^i-7\sum_{n=0}^\infty a_i x^{i+2}+6\sum_{n=0}^\infty a_i x^{i+3}$$ So, for the constant term $$8=a_0$$ For the first power of ... 1 I'm not sure what you are allowed to assume. Do you know that$e$is transcendental and can you use that? If so consider$e^{\frac{p}{q}}=l$where$l\in \mathbb{Q}$. What happens when you look at the polynomial$x^p-l^q$? Looking at the first part of your question I'm assuming it's unlikely you are allowed to use the fact that$e $is transcendental ... 1 As noted elsewhere in the thread, if$f$is irreducible with roots$\{a_1, a_2, ..., a_n\}$, it is definitely not the case that$\mathbb{Q}[x]/\langle f(x) \rangle \cong \mathbb{Q}[a_1, a_2, ..., a_n]$. It is the case, however, that$\mathbb{Q}[x]/ \langle f(x) \rangle \cong \mathbb{Q}[a_j]$for any$1 \leq j \leq n$. More generally, this result holds for ... 1 If$a = 0$, then we have$\lvert f(w)\rvert = \lvert b\rvert$for all$w$, so the strict inequality cannot be achieved. If$a \neq 0$, we can divide by$a$and assume$a = 1$. If$n \geqslant 3$, then $$\{ w^n : \lvert w\rvert = r, \lvert w-1\rvert < 1\} = \{ z : \lvert z\rvert = r^n\},$$ since for every$r < 1$the angle of the arc$\{ w : \lvert ...

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