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The linear algebraic origins of the question are something of a red herring: This may be taken as a question about the derivative of a complex polynomial $p$ at a root $a$. Write $$p(x) = (x - a)^{k} q(x),\qquad q(a) \neq 0.$$ If $k > 1$, then $p'(a) = 0$, regardless of the other roots. If $k = 1$, i.e., $a$ is a simple root of $p$, ...

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GF$(256)$ is small enough that you should construct an antilog table for it and save it for later reference rather than compute the polynomial form of $\alpha^{32}$ or $\alpha^{100}$ on the fly each time you need it. The computer version of the antilog table is an array that stores the polynomial forms for $1 (= \alpha^0), \alpha, \alpha^2, \cdots, \alpha^{... 0 I can tell you how to factorise a Cubic polynomial. This would be a long lecture, so after reading this you try out with some polynomials. Let's Start: A third degree Polynomial is in the form of $$x^3 + bx^2+cx+d$$ Let the roots be$\alpha,\beta,\gamma$Do you Know the symmetric notation: $$x^3 + (\sum_{}^{} \alpha )x^2 + (\sum_{}^{} \alpha\beta )x + ... 0 Hi I have found something new!! I am going to tell you a new formula: Any Quadratic Polynomial ax^2 + bx + c = 0 , whose roots are \alpha and \beta, You just find$$D = (am)^2$$for some complex m. Then put m in this equation$$\alpha = (\frac{-b}{a} + m)\frac{1}{2}\beta = (\frac{-b}{a} - m)\frac{1}{2}$$You will get your ... 1 Hint \ Just like the derivative, the linear operator \, D f(x) = f(x\!+\!1) - f(x) \, acts on polynomials by decreasing the degree by 1,\, since \,D x^n = (x+1)^n-x^n = c_n x^{n-1} +g(x)\, with \,c_n\neq 0,\,\deg g < n-1. For any such linear operator one can solve equations of the form \, D f(x) = g(x)\, for given polynomial g by using ... 1 A solution different from the (excellent) solution of Batominovski (in characteristic zero). Let E_n= the space of the polynomials of degree \leq n. Choose n> the degree of P. Then T(Q)=Q(x+1)-Q(x) is a linear surjective application from E_n to E_{n-1}, as the image of the polynomial x^m for m\geq 1 is U_m=(x+1)^m-x^m=mx^{m-1}+... and ... 10 I am assuming that you are working on \mathbb{R}, \mathbb{C}, or any field of characteristic 0. Write$$P(x)+1=\sum_{r=0}^k\,c_r\,\binom{x}{r}$$for some nonnegative integer k and for some constants c_0,c_1,\ldots,c_k. (These constants exist because \binom{x}{r} for r=0,1,2,\ldots span the vector space of polynomials in x.) Show that$$F(x)... 0 This is basically the inversion for formal power series: If you have a power series $$\sum_i c_i x^i$$ and want to write its inverse as $$\sum_i b_i x^i = \frac{1}{\sum_i c_i x^i}$$ you see that it has to hold $$\sum_i b_i x^i \sum_i c_ix^i \equiv 1.$$ Using the Cauchy product on the left gives some $$\sum_i d_i x^i\equiv 1$$ and comparing of the ... 1 According to the linked paper On Solvable Quintics$X^5+aX+b$and$X^5+aX^2+b$by Blair K. Spearman and Kenneth S. Williams, the solutions are as follows (we take$s=5$in the first equation and$s=1$for the rest): $$x=\omega^{j}u_1+\omega^{2j}u_2+\omega^{3j}u_3+\omega^{4j}u_4$$$\omega- the fifth root of unity. $$x^5-5x^2-3=0$$ $$u_1=\left(-... 1 Guessing the solution can be hard if you're not used to it. Here's something to help you understand how SiongthyeGoh found it. Let f be defined by :$$ \forall x \in \mathbb{R}, f(x) = x^2-3x $$We want to find every x so that f(x)=4. Let's solve it !$$ \begin{align} x^2-3x &= 4& x^2-3x ~~\text{ begins like }~~ x^2-3x+\frac94=\left(x-\... 0 $$x(x-3)=4$$ $$x^2-3x-4=0$$ $$(x-4)(x+1)=0$$ Are you able to complete the rest? Extra comment: In your working, fromx(x-3)=4$, we cannot conclude that$(x-3)=4$. Extra comment 2: note that $$(x-a)(x-b)=x^2-(a+b)x+ab$$ Hence to recover$a$and$b$, I tried for factor of$-4$where the sum is equal to$3$. For general quadractic equation, you can ... 0$x^2-3x=4$can be solved by using the pq-formula :$x^2-3x-4=0x_1=\frac{3}{2}+\sqrt{\frac{9}{4}+4}=4x_2=\frac{3}{2}-\sqrt{\frac{9}{4}+4}=-1$0 Arithmetic with polynomials — including things like divisibility and modular arithmitic — is closely analogous to arithmetic with integers, and many of the same methods apply. In particular, consider the standard techniques for doing similar computations in the integers, e.g. as described at How do I compute$a^b\,\bmod c$by hand?. 0 NOTE: None of the following Python code has been tested. Furthermore, it's not real code but rather pseudo-code adapted to Python in order to explain a rigorous algorithm while making it readable as possible to people who don't know code or don't know Python. Wikipedia has a very helpful algorithm for this and it is about the exact field you are dealing ... 1 My question is now: Does this function converge? Or does this polynomial interpolation somewhen starts flipping and going crazy? It would if the polynomial$f_{m+1}$of interval$[n+1,n+2]$happened to be The Taylor polynomial$T_{m+1}(x)$of degree$m+1$of$f$and the polynomial$f_m$of interval$[n,n+1]$happened to be The Taylor polynomial$T_m(x)$... 1 It is not necessary to shift$X$by$1/2$to symmetrize the polynomial. Any polynomial function of$X(1-X)$is a solution, and all solutions can be written in that form by starting from the highest degree (which must be even) term and working downward. The advantage is that this works "over$\mathbb{Z}$" without any powers of$\frac{1}{2}$. For any ring$...

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Clearly, the set of real polynomials (with coefficient-wise addition and scalar multiplication) forms a vector space over $\Bbb R$: Sums and multiples of polynomials are again polynomials. This is $(1)$. Every polynomial, i.e., every formal(!) expression of the form $a_0+a_1X+\ldots +a_nX^n$ for suitable $n\in\Bbb N$ and $a_0,\ldots,a_n\in\Bbb R$ can be ...

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Some examples: the set of solutions of equation of the harmonic oscillator $\,y''+\omega^2y=0$, is a real vector space of dimension $2$, with basis the functions $\{\cos \omega x,\sin\omega x\}$. The set of continuous functions defined on $[0,1]$ is a vector space. It is not finite dimensional (as most function spaces) The set of polynomials with ...

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$\mathscr{P(\mathbb{R})}$ is a subset of $\mathbb{R}^\mathbb{R}$, and it is itself a vector space. Therefore (in other words), $\mathscr{P(\mathbb{R})}$ is a subspace of $\mathbb{R}^\mathbb{R}$, since $\mathbb{R}^\mathbb{R}$ is also a vector space.

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The general case of factoring a polynomial of degree 3 is quite painful. But in cases encountered in homework/assignements, you can usually: find an obvious root (try 0, 1, -1, i, -i) recognize some patterns (see for instance this example) use wolfram alpha

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I will try our the factor of $2$. Since $2^3-3(2)-2=0$, $x-2$ is a factor. Since $(-1)^3-3(-1)-2=0$, $x+1$ is another factor. Hence I will think of factorizing it as $$x^3-3x-2=(x+1)(x-2)(x+c)$$ $$-2=1(-2)c$$ Hence $c=1$ $$x^3-3x-2=(x+1)^2(x-2)$$

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There is no easy general method to factorize a third degree polynomial. However in your case, you can notice that $2$ is a root of your polynomial : $$2^3-3\times 2-2=0.$$ So you get $$X^3-3X-2=(X-2)(aX^2+bX+c),$$ you develop and by identification you get $$X^3-3X-2=(X-2)(X+1)^2.$$

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Try to "guess" some rational root $\;\cfrac rs\;$ , which by the Rational Root Theorem must fulfill $\;r\,\mid\,-2\;,\;\;s\,\mid\,1\;$ , and indeed $\;2\;$ is a root, so divide by $\;x-2\;$ : $$x^3-3x-2=(x-2)(x^2+2x+1)=(x-2)(x+1)^2$$ and you have one simple root and one double one. If there is no rational root then the task is much, but really much harder ...

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Since you tagged it elementary number theory and not abstract algebra, here is an approach that does not require any knowledge of ring theory. Write the polynomials as $\,A,B.\,$ Suppose $\,p\nmid A,B.\,$ So when reduced mod $\,p,\,$ both are $\not\equiv 0$ hence they have leading coef's $\,a,b\not\equiv 0\pmod p.\,$ By $\,p\,$ prime: $\,p\nmid a,b\,\... 1 Conditioned on the number of roots being$k$, the roots are uniformly distributed over all multi-sets of$\mathbb{F}_p$of size$k$. Let$P(x)$be a uniformly chosen polynomial chosen from $$\mathbb{P} := \{f(x) = x^d + a_1x^{d-1}+\ldots + a_{p-1} x + a_p: a_i \in \mathbb{F}_p, f(x)\text{ has at least one root in }\mathbb{F}_p\}.$$ Each polynomial of$\...

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If $t$ is a root, then we have $p=|p|=|a_1t+a_2t^2+...+t^n|\leq|a_1t|+|a_2t^2|+...+|t^n|$. If $|t|\leq 1$, then $|t^n|\leq 1$ for every $n$, and we have $|a_1t|+|a_2t^2|+...+|t^n|\leq|a_1|+...+|a_n|<p$ and that's a contradiction.

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Another way to think the problem is the following. First, show that $A_5$ cannot have a subgroup of size $20.(1)$ If such subgroup $H$ existed, $H$ would be a subgroup of $A_5$ of index $3$, which cannot be normal, why would this imply $A_5$ to have a normal subgroup of index $2$?; consider the action of $G$ on the cosets of $H$. Let $F$ be the splitting ...

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A polynomial in $n$ variables is symmetric exactly when you can do any permutation on the variables and leave the polynomials unchanged. In other words, you must have $$f(x_1, \dots, x_n) = f(x_{\sigma(1)}, \dots, x_{\sigma(n)})$$ for any permutation $\sigma \in \mathfrak{S}_n$. The standard symmetric polynomial verify this, so a linear combination of them ...

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Think it in the reverse way; Say your field contains $n$ elements. Choose uniformly any number of value from the field, say $p_1,p_2,\cdots , p_k|k\le n$ It is sure that $P(x)=\prod\limits_{i=1}^k(x-p_i)\in\Bbb{F}[x]$

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The roots of any (irreducible) solvable quintic can be found, using methods due to George Paxton Young in 1888. An explicit 3-page formula based on those methods was given by Daniel Lazard in 2004. Source: https://en.wikipedia.org/wiki/Quintic_function

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For the first part let $f$ and $g$ be primitive polynomials in $\mathbb{Z}$ and assume their product isn't, i.e. there exists prime number $p$ that divides all coefficients of $f\cdot g$. Now define the following mapping: $$\phi = \begin{cases} \mathbb{Z}[x] \to \mathbb{Z}_p[x] \\ a_nx^n + a_{n-1}x^{n-1} + ... a_0 \to [a_n]x^n + [a_{n-1}]x^{n-1} + ... + [... 1 For any 4\times 4 invertible matrix A the polynomials$$ \begin{bmatrix} p_1(t)\\p_2(t)\\p_3(t)\\p_4(t) \end{bmatrix}=A\begin{bmatrix} 1\\t\\t^2\\t^3 \end{bmatrix} $$constitute a basis. 1 Since this vector space K[t]_3 is isomorphic to K^4, where K is the field you are taking polynomials over, there is a one-to-one correspondence between bases of K[t]_3 and invertible 4\times 4 matrices over K, with the entries in column k giving coefficients of the k-th basis polynomial. 1 Any (P_0, P_1, P_2, P_3) with \deg P_i=i will be a basis of polynomials of degree three or less. Indeed, it is easy to show that such polynomials are linearly independent. 0 An iterative method for finding the best "maximum norm" approximation by polynomial of degree at most d to a given smooth function f(x) on a bounded interval I = [a,b] was proposed by Evgeny Yakovlevich Remez in 1934, and has come to be known as the Remez exchange algorithm. The underlying idea is a characterization of the "mini-max" polynomial ... 1 Hint   We do a slightly less trivial case \,a=\sqrt2,\,b=\sqrt[3]3.\, \Bbb Q(a,b)\cong \Bbb Q\langle1,b,b^2, a,ab,ab^2\rangle as a vector space. \,x\mapsto (a\!+\!b)\,x\, is a linear map on this vector space. Compute its matrix, then apply Cayley-Hamilton, to compute a (characteristic) polynomial having \,a\!+\!b\, as root (e.g. by computing a ... 0 We have \alpha=\sqrt{2}+\sqrt{3} \in K=\mathbb Q(\sqrt2, \sqrt3). \mathcal B = \{1, \sqrt2, \sqrt3, \sqrt6\} is a basis for K over \mathbb Q. Write the matrix of \mathbb Q-linear map \mu : x \mapsto \alpha x with respect to \mathcal B. Then \alpha is a root of the characteristic polynomial of \mu because of the Cayley–Hamilton theorem, ... 1 Gaussian elimination is the key. The ring$$ \mathbb{Q}[x,y]/(P(x),Q(y)) $$is a vector space over \mathbb{Q} with dimension \partial P\cdot \partial Q, and a base given by x^i y^j for i\in[0,\partial P-1],j\in[0,\partial Q-1]. By representing 1,(x+y),(x+y)^2,\ldots,(x+y)^{\partial P\cdot\partial Q} in such a base we get \partial P\cdot\partial Q+... 0 Daniel Lazard implemented a program in Maple that solves for quintics that are solvable by radicals: you can take a look at this book published by Springer (2002): The Legacy of Niels-Henrik Abel 1 We just have to find the minimal polynomial of$$ \alpha = \frac{\cos\frac{19\pi}{22}}{\cos\frac{\pi}{22}}=-\frac{\cos\frac{3\pi}{22}}{\cos\frac{\pi}{22}}=3-4\cos^2\frac{\pi}{22}=1-2\cos\frac{\pi}{11} \tag{1}$$then find the conjugate roots. But it is well-known that the minimal polynomial of \cos\frac{2\pi}{m} has degree \frac{\varphi(m)}{2} (i.e. 5 ... 7 Using the equality \cos(\pi \pm x)=-\cos(x) you get$$x_1=-\frac{\cos \frac{3 }{22} \pi}{\cos \frac{1}{22} \pi} \\ x_2=-\frac{\cos \frac{9}{22} \pi}{\cos \frac{3}{22} \pi} \\ x_3=-\frac{\cos \frac{15}{22} \pi}{\cos \frac{5}{22} \pi} \\ x_4=-\frac{\cos \frac{21}{22} \pi}{\cos \frac{7}{22} \pi} \\ x_5=-\frac{\cos \frac{27}{22} \pi}{\cos \frac{9}{22} \pi} ... 0 The answer in the linked question still applies. Consider the matrix \begin{align*} y &= \begin{pmatrix} 0 & \cdots & 0 & 1 \\ 0 & \cdots & 0 & 0\\ \vdots & \ddots & 0 & 0\\ 0 & \cdots & 0 & 0 \end{pmatrix} \in M_n(\mathbb{R}). \end{align*} If y = x^n for some x\in M_n(\mathbb{R}), then all the ... 1 There is no need to consider eigenvalues or eigenvectors. You have: A = B^{-1} D B $$with D being a diagonal matrix. It follows that for any n\in\mathbb{N}$$ A^n = (B^{-1}DB)\cdot(B^{-1}DB)\cdot\ldots\cdot(B^{-1}DB)= B^{-1} D^n B$$holds, so for every polynomial p\in\mathbb{R}[x]$$ p(A) = B^{-1} p(D) B$$holds too. Since p(D) is a diagonal ... 3 It is false. Consider the matrix.$$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$It is not the square of a complex matrix, so p(X)=X^2 is not surjective. 1 We can write p into this form:$$p(x) = a_mx^m + a_{m−1}x^{m−1} + · · · + a_1x + a_0$$Now let \Bbb v be an eigenvector of A with eigenvalue \lambda. Since A^k\Bbb v = \lambda ^k\Bbb v for every k, we see that$$p(A)v = a_mA^m\Bbb v + a_{m−1}A^{m−1}\Bbb v + · · · + a_1A\Bbb v + a_0I\Bbb v \\= a_mλ^m\Bbb v + a_{m−1}λ^{m−1}\Bbb v + · · · + a_1λ\Bbb v ...

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I'll try to address some of the misunderstandings I think you have based on the question and the comments. Someone else can tackle the problem given in $(3)$ of explaining everything about Taylor polynomials and Taylor series from the ground up if they want (though I doubt anyone will). First off, it sounds like you're conflating Taylor polynomials with ...

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$x = p(1 + (\frac qp)^{\frac 15})^5 \in \Bbb Q((\frac qp)^{\frac 15})$. And so the splitting field of your degree $5$ polynomial is $\Bbb Q((\frac qp)^{\frac 15},\zeta_5)$, the conjugates of $x$ are obtained by multiplying $(\frac qp)^{\frac 15}$ with a power of $\zeta_5$. Also this is not really a degree $25$ equation. If you naïvely look at the "...

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$s^4 + 14s^3 +45s^2 +650s + 1800$ is irreducible over $\mathbb Q$ because it is irreducible mod $11$. In particular, its roots are irrational numbers. There is a formula for the roots but it'll probably be very ugly.

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This is false. For example, $X^2+2$ and $X^2+6$ are both equal and irreducible mod $4$. However, since $\mathbb Q_2$ does not contain a square root of $3$, their roots give different extensions of $\mathbb Q_2$.

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Hint: You have a product of two polynomials. Crop the problem into smaller problems in finding the zeros of each polynomial. For $z^2+iz+2=0$: Complete the square and solve for z or simply use the quadratic formula: $z_{1/2}=\frac{-i\pm\sqrt{i^2-4\cdot \cdot 2}}{2\cdot 1}=\frac{-i\pm\sqrt{-9}}{2\cdot 1}=\frac{-i\pm 3i}{2\cdot 1}$ For $z^3-8i=0$: Rewrite \$...

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