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The theorem may as well state that every polynomial equation of degree $n$ has exactly $n$ roots (counted with their multiplicity). The statements are equivalent, for, if your polynomial $p(z)$ of degree $n$ has one root $\lambda$, then you can factor it as $$p(z) = (z-\lambda)q(z),$$ where the degree of $q$ is $n-1$ Then you can recursively apply the ...

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This is a statement of the Fundamental Theorem of Algebra that is completely correct, but in a way kind of obscures what is actually going on if someone doesn't think about what this actually means... Consider a complex polynomial $p(z) = a_0 + a_1z + ... + a_nz^n$. By the fundamental theorem of algebra, this has at least 1 zero. It is a well-known result ...

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Suppose that all the roots are non zero. Put $x_1+x_2=u$, $a=x_1x_2$, $x_3+x_4=v$, $b=x_3x_4$, we suppose that $u\in\mathbb{Q}$, then it is also the case for $x_3+x_4$, as the sum of all the roots is rational, and that $a,b$ are irrationals (as the product $ab$ is rational, if $a$ is irrational, then it is also the case for $b$). The polynomial with roots $... 7 Note that your polynomial is a geometric series with first term$-x$, common ratio$-x^2$so that it can be written as $$\frac{x(x^{16} - 1)}{x^2 + 1} = 7$$ So $$x^{16} = \frac{7(x^2 + 1)}{x} + 1$$ And now, all you need to do is show that$\frac{x^2 + 1}{x} = x + \frac{1}{x} > 2$, which is not hard. 6 It can't be done. There are formulas for the roots of a quadratic, cubic or quartic in terms of radicals, but not (in general) for the roots of a polynomial of degree$5$or higher. For example, the roots of$x^5 + 2 x + 1can't be written in terms of radicals. See e.g. Abel-Ruffini theorem 5 Here's a good translation. Gauss' original writing is a little old fashioned but still perfectly readable. http://archive.larouchepac.com/node/12482 The flaw doesn't have anything to do with hypergeometric functions as far as I know. The trouble is he assumes without proof that a real algebraic curve that enters a circle must leave it again. 5 An efficient algorithm to factor a polynomial into irreducible polynomials is given in this article. The lattice basis reduction algorithm they developed for this purpose is the famous LLL algorithm which has many applications besides its use in polynomial factorization problems. 5 Your doing is correct, you just need to expand the expressions inside the parenthesis: \begin{align}((x-1) - (x-5))((x-1)+(x-5))&=(x-1-x+5)(x-1+x-5)\\ & =4(2x-6)\\ & =8(x-3)\end{align} 4 If\lambda_1 = re^{i\theta}$,$\lambda_2 = re^{-i\theta}$, then $$\begin{cases} a = \lambda_1 + \lambda_2 = 2r\cos\theta\\ b = \lambda_2 \lambda_2 = r^2 \end{cases} \quad\implies\quad \begin{cases} r &= \sqrt{b}\\ \cos\theta &= \frac{a}{2\sqrt{b}} \end{cases}$$ For integer$M > 1$, $$P_M = \frac{\lambda_1^M - \lambda_2^M}{\lambda_1 - \lambda_2}... 4 After your findings so far, it suffices to show that the derivative$$ f'(x)=4x^3+4x-6$$has only one real root. For this again, it is sufficient to observe that the second derivative$$ f''(x)=12x^2+4$$is strictly positive. In other words, your Rolle aproach should work: If there were three real roots x_1<x_2<x_3 of f, we'd have roots \xi_1,\... 3 Let f(x) be the left side. Note that f(x) \to 0 as x \to \pm \infty, +\infty as x \to a_i- and -\infty as x \to a_i+. So there will be at least one root in each interval (-\infty, a_1), (a_1, a_2), ..., (a_{n-1}, a_n). Since the equation is equivalent to a polynomial equation with degree n, there are at most n roots. Therefore ... 3 Don't know if this is the quickest/easiest/best/etc. way but here's one way.. P has two inflection points, which means P''(x) has two distinct real zeros. Therefore P''(x) has degree at least 2, which means P(x) has degree at least 4. Since P(x) is a polynomial of least degree, we can conclude that P(x) must have degree 4. Therefore we can ... 3 2y^4 - 4y^3 + 2y^2 + 1 =0 is nothing but 2y^2\cdot (y-1)^2 + 1 = 0 which is sum of square of two non negative terms. 3 Using a Pell-like equation, there are in fact infinitely many positive integer solutions to,$$xyzw=504(x^2+y^2+z^2+w^2)\tag1$$1st family:$$\big(x,y,z,w\big)=\big(84,\;84,\;21q,\;21(4p+7q)\big)$$where,$$p^2-3q^2=-2\tag2$$An initial point is (p,q) = (-1,1) yielding the OP's known \big(x,y,z,w\big)=\big(84,\;84,\;21,\;63\big). As (2) has ... 3 The simple answer, in the spirit of the comments, is that all polynomials are functions but not all functions are polynomials. A function is simply a rule that assigns a value in the codomain to every value in the domain. A couple simple examples of functions that are not polynomials are \sin x and |x|. A less simple one is the Dirichlet function which ... 3 Since x_1x_2 is irrational, there is an automorphism \sigma of \overline {\Bbb Q} that changes x_1x_2 into something else. Since \sigma acts on a permutation on the roots, we must have \sigma(x_1)=x_i and \sigma(x_2) = x_j where i,j \in \{1;2;3;4\} and i \neq j, and importantly, x_1x_2 \neq x_ix_j Since x_1+x_2 is rational it is fixed ... 3$${ \left( x-1 \right) }^{ 2 }-{ \left( x-5 \right) }^{ 2 }=\left( x-1-x+5 \right) \left( x-1+x-5 \right) =4\left( 2x-6 \right) =8\left( x-3 \right) $$2 Solution credits to Rui Yao: According to a property of finite differences, if we set c_n\in \mathbb{Z} to be the highest coefficient of f(x), which has degree n, then$$n!c_n=\Delta ^n [f](x)=\sum_{i=0}^{n}\binom{n}{i} f(i)(-1)^{n-i}$$Thus$$|n!c_n|=|\sum_{i=0}^{n}\binom{n}{i} f(i)(-1)^{n-i}|\le \sum_{i=0}^{n}|\binom{n}{i} f(i)(-1)^{n-i}|\... 2 Let$f(x)$be a polynomial. Suppose that the distinct primes$p$and$q$are roots of$f(x)\equiv 0\pmod{pq}$. Since$(x-p)(x-q)$is monic, there is a polynomial$g(x)$such that$f(x)=(x-p)(x-q)g(x)+ax+b$for some integers$a$and$b$. Since$p$and$q$are roots of$f(x)\equiv 0\pmod{pq}$, it follows that$pq$divides$ap+b$and$pq$divides$aq+b$. So ... 2 (High School Level) GUESS AND CHECK METHOD: This method works well if you are just beginning and before too long you will start to see patterns and factor quickly. This speed comes in handy if you need to factor more than one expression in the same question or if you need to do a long list of factoring questions. (1) GUESS: First, think of two possible ... 2 Or more simply, as you mentioned as an easy proof, one has $$\frac{\lambda_1^M - \lambda_2^M}{\lambda_1 - \lambda_2} = \sum_{k=0}^{M-1} \lambda_1^{M-1-k} \lambda_2^{k}$$ Hence $$\bbox[lightgreen,5px,border:2px solid green]{P_M(a,b) = \frac{1}{2^{M-1}}\sum_{k=0}^{M-1} \left(-a + \sqrt{a^2 - 4b}\right)^{M-1-k} \left(-a - \sqrt{a^2 - 4b}\right)^{k}}.$$ 2 In my opinion this is a bit simpler to prove, if we interpret the matrix$P_n$as a linear transformation. Consider the space$V_n$of polynomials of degree$\le n+1$(over, say$\Bbb{Q}$, but you are welcome to use reals or complex numbers, or any other field actually). The mapping$T: f(x)\mapsto f(x+1)$for all$f(x)\in V_n$is obviously linear. My key ... 2 The MacTutor biography mentions "Disquisitiones generales circa seriem infinitam" as introducing the hypergeometric functions. This was not his doctoral thesis: it is dated 1812. His doctoral dissertation "Demonstratio Nova Theorematis Omnem Fvnctionem Algebraicam Rationalem Integram Vnus Variabilis In Factores Reales Primi Vel Secvndi Gradvs Resolvi Posse"... 2 Notice that the left hand is equal to$(x^3-x)(x^{16}-1)/(x^4-1)$; hence the equation will be transfer into:$(x^{16}-1)=7(x^4-1)/(x^3-x)=7(x^2+1)/x$; notice that$(x^2+1)/x$is graeter than 2, hence the right hand is greater than 7*2, therefor$x^{16}>15$. 2$ x^{14} - x^{12} + \ldots + x^2 -1 = \frac{7}{x}, $I am considering$ x \neq 0 $multiply both side by$ x^2 $and add you get$ x^{16} = 1+ 7x + \frac{7}{x}. $Take minimum of R.H.S. 2 This is a polynomial of degree$n-1$in$x$(unless$a=n$). If, for example,$a$is an integer and$p$is a prime such that$p$but not$p^2$divides$a$while$p$does not divide$n$, then Eisenstein's criterion says it is irreducible over the rationals. In general, for$n \ge 6$we would expect it not to have roots expressible in radicals. EDIT: For ... 2 A well known result is that any integer-valued polynomial is the sum of integer multiples of$\binom{x}{k} =\dfrac{x(x-1)...(x-k+1)}{k!} $. https://en.wikipedia.org/wiki/Integer-valued_polynomial By multiplying by the largest$k!$, we get your statement. 2 It is the remainder theorem that if you divide$P(x)$by the degree two$x^2-3x+2$, the remainder will be a degree one polynomial. Hence, $$P(x) = (x^2-3x+2)Q(x) + (ax+b)$$ Now, this is an identity for all values of$x$. Thus if in particular you plug in$x=1$on both sides, you'll get an equation. But notice that$x=1 \implies x^2-3x+2=0$, so you're left ... 2 I don't quite understand the work you showed without any formatting (not sure where one statement begins and the next ends). I'll write out how I approached the problem. We know that for some polynomials$f, g$$$P(x) = (x-1)f(x) + 5$$ $$P(x) = (x-2)g(x) + 3$$ Multiplying$(x-2)$to the top equation and$(x-1)$to the bottom gives $$(x-2)P(x) = (x-1)(x-2)... 2 using the formula$$a^2-b^2=(a-b)(a+b)$$we obtain$$(x-1)^2-(x-5)^2=(x-1-x+5)(x-1+x-5)=4(2x-6)=8(x-3)$\$

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