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How you can go about trying to understand this: Note that the given equation is a quadratic equation in $y$, whose coefficients are polynomials in $x$. That is, we can rewrite it as $$y^2 + by + c = 0$$ where $b = 3x$ and $c = -5x^3 + 1$. Thus using the quadratic formula as usual we find that $$y = \frac{-3x \pm \sqrt{9x^2 + 20x^3 - 4}}{2}$$ which is ...

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It doesn't define $y$ as a function of $x$, because it doesn't pass the Vertical Line Test. Take for example $x=1$, which gives $y^2+3y-4=0 \Leftrightarrow y=1 \vee y=-4$.

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Note that $\beta^4 + \gamma^4 + \delta^4 = S_4 - \alpha^4$, and similarly for the other quantities in parentheses. Substitute this in your desired expression, and expand it. You can then write the entire expression in terms of various $S$'s.

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Hint: Consider any run of $n$ consecutive integers, say: $$a + 1, a + 2, \ldots, a + n$$ Summing them together, we have: $$\sum_{k=1}^n (a + k) = \sum_{k=1}^n a + \sum_{k=1}^n k = an + \frac{n(n + 1)}{2} = n\left( a + \frac{n + 1}{2}\right)$$ Now for what values of $n$ will the expression in brackets be an integer?

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To compute the polynomial using a brute force approach, you can always use this : http://en.wikipedia.org/wiki/Lagrange_polynomial Your polynomial has degree $4$, so it will be a bit painful but you can expand it using a computer program if you're not up for it. Alternatively, the standard approach to compute this polynomial is using telescopic sums. ...

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The question in your title is different than the question you ask at the end of your post. So I will answer them both. Firstly, it's not true that polynomials with rational coefficients have integer roots. For instance, $x^2 - 2$ has roots $\pm \sqrt 2$. However, it is true that polynomials of the form $p(x) = a_0 + a_1 {X \choose 1} + \ldots + a_d {X ... 1 Hint: compute $$P(x+1) - P(x)$$ with$P(x) = ax^4 + bx^3 + cx^2 + dx$and try to find$a,b,c,d$so that$P(x+1) - P(x) = x^3$0 Hint Note that$27^y=(3^y)^3$,$9^y= (3^y)^2$, if$x=3^y$then: $$12 \cdot 27^y + 25 \cdot 9^y -4 \cdot 3^y -12=0\equiv 12x^3+25x^2 -4x -12=0$$ 0 $$27^y = (3^3)^y = 3^{3 \cdot y} = 3^{y \cdot 3} = (3^y)^3$$ 2 Use Euclid's algorithm in the usual way. The first step is $$(x^6+x^5+x^4+x^3+x^2+1)=(x^2)(x^4+x^3+x^2+1)+(x^3+1)$$ and the next will look like $$(x^4+x^3+x^2+1)=(\cdots)(x^3+1)+(\cdots)\ .$$ 0 it is$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2=(x^2+1-\sqrt{2}x)(x^2+1+\sqrt{2}x)$1 It is the ideal generated by$\{x,y\}$. Explicitly,$(x,y)$consists of those polynomials whose constant term is zero. Hint for the exercise: If$(x,y)=(g)$, then show that$g$is a gcd of$x$and$y$. But$x$and$y$are coprime. 1 Given a multivariate polynomial, a good place to start would be to factor out$x$from all the terms with$x$, then factor$y$out of all the terms that are left with$y$, and so on. Given a random multivariate polynomial of degree$n$in$m$variables, on average$O(n^m)$terms will have an$x$, meaning just factoring out an$x$will reduce the number of ... 0 Hint: the first condition gives$(x+1)^2$as a factor, while the second gives$x-3$. So we must have$f(x)=a(x+1)^2(x-3)$for some non-negative constant$a$. Use the remainder theorem to find$a$. 0 Any quartic where terms of odd degree have zero coefficient,namely any quartic polynomial of form$f(x)=x^4+x^2+n$where$n$is any non-negative integer 0 So I think I found the answer using the conditition$gcd(r,p)=1$, which I didn't previously mention. It would be nice if someone could verify this answer. Since we are in$\mathbb{Z}[X]/(X^r-1,p)$,$X$is a root of unity. But even better, since$gcd(r,p)=1$,$X^p$is a root of$X$.$\implies \exists h : X^{ph} \equiv X$. From Edit3 we have ... 1 Hint: Substitute$q^5$with$r$, so you get the equation$r^2-2r+2=0$3 You have to solve it in two steps. Let$x=q^{5}$. Then$x²-2x+2=0$Solve this, and then solve the first equation to find the solutions in$q$. 1 Step ii was done by long division: the polynomial$x^4 + 9x^3 + \ldots$was divided by$x^2 + 10 x + 41$to get$x^2 - x + 4$, with a remainder of$-160$. 2 It seems to be they simply divided with residue the two polynomials: $$x^4+9x^3+35x^2-x+4=(x^2+10x+41)(x^2-x+4)-160$$ 1 Suppose you have$p(x)=(x-b)f(x)$for some polynomial$f(x)$... Product of constant term of$f$and$-b$should be same as constant term of$p$. Thus,$b$should divide constant term... So, it is enough to check for factors of$p(0)$.. 1 But this require to check every b∈ℤ where ℤ is an infinite set. False. Only is required check the divisors ($+$and$-$) of$p(0)$. 1 Suggestion. Instead of doing what you did for $$f(\vartheta)=\frac{1}{1+\sin^2\vartheta},$$ try in on each of the two terms of what$f(\vartheta)$is equal to: $$\frac{1}{1+\sin^2\vartheta}=\frac{i}{2}\left(\frac{1}{i+\sin\vartheta}+\frac{1}{i-\sin\vartheta}\right).$$ 1$\int^{\pi}_0 \frac{1}{1+ \sin^2 \theta}$d$\theta =\int^{\pi}_0\frac{1}{1+\frac{1}{2}-\frac{\cos 2\theta}{2}}$d$\theta = \int^{\pi}_0\frac{1}{\frac{3}{2}-\frac{\cos 2\theta}{2}}$d$\theta = \int^{\pi}_0\frac{2}{3- \cos 2 \theta}$d$\theta$I believe that this can then be computed using Weierstrass substitution, and forming an improper integral. ... 0 All you know is this: a nonzero polynomial of degree$d$cannot have more than$d$roots in some (extension) field. 2 No. Take$p(X)=X(X+1)$,$F=\Bbb F_2$,$K=\Bbb F_4$. Indeed, let$\beta\in\Bbb F_4-\Bbb F_2$. We know that$\beta^2+\beta=1$, that is,$p(\beta)=1$. And$p(0)=p(1)=0$. 1 Hint: $$x^{n+1} - 1 = x(x^n - 1) + (x - 1)=x(x-1)(?) + (x - 1)=(x-1)(x(?) + 1)$$ Alternative: Actually We don't need to use Induction, Let$f(x)=x^n-1$Now$f(1)=0$, That means$1$is a root if$f(x)$so$f(x)=(x-1)\times(?)$( By '$?$' I mean some another factors as$f(x)$is polynomial of degree$n$) 0 Yes, it is true that all unlinks have Alexander-Conway polynomial equal to zero. This comes from the more general fact that the Alexander polynomial of a splittable link is always 0. A link is splittable if its components can be seperated by a plane in$\mathbb{R}^3$, which is exactly what you would want it to mean. And the unlink obviously falls into ... 1 To find the product of the roots of a polynomial use Vieta's formula which says if$\lbrace r_n \rbrace$is the set of roots of an$n^{th}$order polynomial$a_nx^n + a_{n-1}x^{n-1} + \dots +a_1x + a_0$, then the product of the roots$r_1r_2 \dots r_n = (-1)^n\frac{a_0}{a_n}$. To find the sum of the roots you use the formula$\sum_{i=0}^{n}r_i = ...

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If it is allowed to use a temporary variable, the number of multiplications can be reduced to 2, for example: $$y=x\cdot x;\quad (y-x-x+5)\cdot y+x-6.$$

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If you do it this way, you only need to do three multiplications. $$x^4 - 2x^3 + 5x^2 + x - 6 = x(x^3 - 2x^2 + 5x + 1) - 6 = x(x(x^2 - 2x + 5)+1)-6=x(x(x(x-2)+5)+1)-6$$

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This is a polynomial of the fourth degree and it has four complex roots. It can be written as $(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ where $x_1 , x_2 ,x_3,x_4$-roots of the polynomial . Here you have 3 multiplication and you can't get less becouuse for getting $x^4$ you need 3 multiplication

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We prove by contradicts: If $p(x)$ have a root $b$ which multiple root occurring more $n-1$ times, then : $$p(b)=p'(b)=\cdots=p^{(n-1)}(b)=0$$ $$b\neq x_1,x_2,\cdots,x_n$$ Denote $q(x)=(x-x_1)^2(x-x_2)^2\cdots(x-x_n)^2$, then as $q(x)\neq0$ in some $\epsilon$-neighborhood of $b$, we consider $\frac{p(x)}{q(x)}=\sum_{i=1}^n\frac{1}{(x-x_i)^2}$: ...

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The first two observations can be explained by the fact that if $r$ is the root of $a_nz^n+\cdots +a_0$ then $r^{-1}$ is the root of $a_0z^n+\cdots +a_n$. Since the joint probability density of the coefficients is symmetric under $a_k\mapsto a_{n-k}$ transformation the density of the roots should be symmetric under $r\mapsto r^{-1}$. It should be fairly ...

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The septic discriminant has 1103 terms. See OEIS sequence A007878 for more.

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If you have a complex root $c_k$ such that $|c_k-i|<1, |=|\overline{c_k-i}|<1$ (modulus of complex number is equal to modulus of its conjugate). So $|\bar{c_k}+i|<1$. But as all coefficients are real, $\bar{c_k}$ is also a root of P(z), by the complex conjugate root theorem. And so, there is at least one root $\alpha$, such that $|\alpha + i|<1$ ...

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Since the other answers have worked directly with powers of $x$, an alternative perspective may be interesting. Let $\tau =\ln x$. Then $x^n+x^{-n}=e^{n\tau}+e^{-n\tau}=2\cosh n\tau$. For $n=1$ in particular we have $2\cosh\tau=x+x^{-1}=5$ by assumption, and so we may write $$x^n+x^{-n}=2\cosh(n\tau)=2\cosh(n\cosh^{-1}(5/2)).$$ So this problem can be ...

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You are correct, without factoring, $\frac{f}{g}$ is a bounded entire function, hence constant by Liouville's theorem.

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there are only two irreducible polynomials of degree $3$ in $F_{2^3}$. these are easy to construct as: $f=x^3 +x^2 +1$ and $g=x^3+x +1$. if $\alpha$ is a root of $f$ then the remaining two roots of $f$ are $\alpha^2$ and $(\alpha^2)^2$ (c.f. Frobenius automorphism). by simple algebra if $\alpha$ is a root of $f$ then $\alpha^{-1}$ is a root of $g$. from ...

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The number of primitive elements in a finite field $GF(n)$ is $φ(n - 1)$, where $φ(m)$ is Euler's totient function. Now compute $\phi(255)$. Use that $255=3\cdot 5\cdot 17$.

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You can find more about this type of equations under the name reciprocal equation or reciprocal polynomial. See, for example, also this post Quadratic substitution question (And several of the posts shown there among linked questions.) In this particular case, you have: \begin{align} x^4+2x^3-14x^2+2x+1&=0\\ x^2+2x-14+\frac2x+\frac1{x^2}&=0 ... 1 The factor theorem states that if x-a is a factor of some polynomial f(x), then f(a)=0. The proof of this is very straightforward. Try it as an exercise. To find factors of your cubic,f(x) = x^3-x^2-14x+24$$We will try to find values of x for which f(x)=0. Supposing that the polynomial can be neatly factored into integer factors x-a, ... 2 We can hope there is a rational root. Real world cubics generally don't have such a root, but school cubics often do. By the Rational Root Theorem, such a root has to be an integer that divides 24. Soon we find that x=2 works. Divide the given cubic polynomial by x-2. We get a quadratic, and you know how to find roots of quadratics. 6 As rogerl said, by symmetry the number of roots inside the unit circle is equal to the number outside. Now, how many are on the unit circle? Let's suppose 1 is not a root, i.e. 2 + 2 b + c \ne 0. The Möbius transformation  w = i(1+z)/(1-z) (z = (w-i)/(w+i)) takes the unit circle (except for the point 1) to the real line, and p(z) = 0 becomes ... 7 p\left(\frac{1}{x}\right) = \frac{1}{x^4}p(x). So unless there are roots on the unit circle (which is not ruled out in the problem as stated), there are two inside and two outside the unit circle. 12 This is an answer for the odd case. Proposition. Let k=1,2,3,\ldots. Then$$ \int_0^1 B_{2k+1}(x)\: \psi (x+1) \:dx=(-1)^{k+1}\frac{(2k+1)!}{(2\pi)^{2k+1}}\pi \: \zeta(2k+1)-\sum_{j=0}^{2k}\!\frac{ {{2k+1}\choose j} B_j}{2k+1-j} \quad (*) $$The proof is here, observing that$$\begin{align} \int_0^1 B_{2k+1}(x)\: \psi (x+1) \:dx & = \int_0^{1} ...

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Take $R=\mathbb Z_6\times\mathbb Z_6\times\cdots$, and $f\in R[X]$, $f(X)=X^3-X$.

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A general strategy to find functions with desired properties is to study a parametrized family of functions. This is exactly the path to the bifurcation diagram in real dynamics and the Mandelbrot set in complex dynamics. A super-attractive orbit of period 3 After a little experimentation, I decided to look at the family of cubic polynomials ...

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For (b), maybe you should give an explicit counterexample: E.g. Let $f(x) = g(x) = x$.

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The trilogarithm has the antiderivative, $$\int\operatorname{Li}_{3}{\left(x\right)}\,\mathrm{d}x=x\operatorname{Li}_{3}{\left(x\right)}-x\operatorname{Li}_{2}{\left(x\right)}+x-x\ln{\left(1-x\right)}+\ln{\left(1-x\right)}+\color{grey}{constant}.$$ So, integrating by parts we find after integrating all the terms with simple antiderivatives: ...

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