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33

Between any two real roots of a polynomial there should be at least one root of its derivative. So the maximum possible of roots in the polynomial is the number of roots of the derivative plus one. In this case, we have $f(x)=x^5+x^4/2+x^3/3+x^2/4+x/24+1/120$, and $$f'''=60x^2+12x+2,$$ which has no real roots. So $f''$ has at most one real root; so $f'$ ...

11

It turns out that any polynomial or rational function that is always positive can be written as a sum of squares. e.g. $$x^4 + x^3 + x^2 + x + 1 = \left(\frac{x^2 + x}{\sqrt{2}}\right)^2 + \left(\frac{x + 1}{\sqrt{2}}\right)^2 + \left(\frac{x^2}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2$$ Alas, I don't know of any systematic way to figure ...

9

Here is a way to do it without resorting to calculus: Eliminate the quartic term (making it into a "depressed quintic") by making the substitution $x=z-1/10$ (as $1/10$ is one fifth of the coefficient of $x^4$), which turns the polynomial into $$z^5+\frac{7 z^3}{30}+\frac{17 z^2}{100}+\frac{z}{6000}+\frac{239}{37500}\text{.}$$ By Descartes' Rule of Signs, ...

7

To give a non-constructive but general algebraic answer: The numbers $\sqrt{2}$ and $\sqrt{7}$ are algebraic. As the algebraic numbers form a field, the number $\sqrt{2}+\sqrt{7}$ is algebraic, too. Thus there is a polynomial with rational coefficients which has $\sqrt{2}+\sqrt{7}$ as root. Now multiply by the common divisor to obtain a polynomial with ...

5

Yes, there is a much simpler method (it boils down to subtracting two linear polynomials) \begin{eqnarray}{\rm mod}\,\ \color{#0a0}{x^2\!+\!x\!-\!3}\!:\ \ f &=& 2x\!-\!1 + (ax\!+\!b)\,(\overbrace{\color{#0a0}{x^2\!+\!x\!-\!3}}^{\large \equiv\ 0\ }\color{#c00}{+1})\\ &\equiv& 2x\!-\!1 + (ax\!+\!b)\,\color{#c00}{(1)}\\ &\equiv& ... 5 If this polynomial were to factor, it would have to factor into a polynomial of degree 2 and a polynomial of degree 1. \begin{align*} x^2 + y^2 + z^2 - xyz - 2 &= \big([\text{terms of degree } 2] + [\text{terms of degree } \le 1] \big) \\ &\quad \big([\text{terms of degree } 1] + [\text{terms of degree } \le 0] \big) \\ \end{align*} The terms of ... 4 We have\sum_{k=0}^n x^k=\frac{x^{n+1}-1}{x-1}$$hence$$\sum_{k=0}^n x^k=\prod_{k=1}^{n}\left(x-e^{2ik\pi/{n+1}}\right)$$so if n is odd say n=2p+1 then ... 3 The standard argument for this requires a bit of machinery, namely the following theorem (the formulation below is from Atiyah-MacDonald, Proposition 5.1). Theorem. Let B be a (commutative) ring and A a subring and x \in B. Then the following four statements are equivalent. x is integral over B; A[x] is a finitely generated A-module; A[x] ... 2 We require that f(x,y)\in\mathbb{Z} for x,y\in\mathbb{Z}. Represent f as a polynomial in y with coefficients in \mathbb{Q}[x]. Since \operatorname{im}f|_{\mathbb{Z}^2}\subseteq\mathbb{Z} we know that f can be made to take the form$$ f(x,y) = \sum_j f_j(x)\binom{y}{j} \quad\text{with}\quad f_j(x)\in\mathbb{Q}[x];\quad ...

2

If you can factor it in $\mathbb{C}$ you already have factored it in $\mathbb{R}$ - just multiply each pair of complex monoids by each other, since each complex root has a conjugate root as well. Since your roots are $(-1)^{k/(n+1)}$, and $(-1)^{k/(n+1)}+(-1)^{(n+1-k)/(n+1)}\in\mathbb{R}$, you should be able to factor out your polynomial. If $n$ is even, ...

2

$f$ satisfies the equation $$f(x)^2 - 25x^4 - 25 = 0$$ which makes it algebraic (not transcendental) by the standard definition given here. Perhaps you are thinking of a different function. Edit: $f(x) = \sqrt[5]{x^4 + 1}$ is also algebraic. In this case, $$f(x)^5 - x^4 - 1 = 0$$

2

Recall what the definition of the resultant of $f$ and $g$ is. That is, if $f = a_nx^n + \cdots a_1x +a_0$ and $g = b_mx^m + \cdots b_1x + b_0$ then $\operatorname{Res}(f,g)$ is the determinant of the Sylvester matrix $\operatorname{Syl}(f,g)$, where $\operatorname{Syl}(f,g)$ is the $(m+n)\times(m+n)$ matrix, $$\operatorname{Syl}(f,g) = \begin{bmatrix} a_0 ... 2 That proof looks good! Having another way to see the result may help your confidence in it: the multiplicative group of nonzero elements of \mathbf{F}_{p^n} is a cyclic group of order p^n - 1. In particular, its order relatively prime to p, and thus every nonzero element has a unique p-th root. (and 0 has a unique p-th root too, obviously) 2 The mere fact that f and g are irreducible in k(x)[y] does not necessarily mean that they are relatively prime. The could still be associates in k(x)[y], say f = y and g = xy. You'll need some machinery, using the assumptions that they're irreducible in k[x,y] and not associates in k[x,y], to argue that this can't happen. It looks like that ... 2$$x^4+ax^2+3=0\\ \implies (x^4+ax^2+3)x=(0)x=0\\ \implies x^5+ax^3+3x=0$$Subtracting from x^5+ax^3+2=0 gives$$x^5+ax^3+2-(x^5+ax^3+3x)=0\\ \implies x^5+ax^3+2-x^5-ax^3-3x=0\\ \implies 2-3x=0$$So,$$x=\dfrac{2}{3}$$Substitute:$$x^4+ax^2+3=0\\ \implies a=-\left(\dfrac{\left(\dfrac{2}{3}\right)^4+3}{\left(\dfrac{2}{3}\right)^2}\right)$$2 For problem (ii), note that \lvert e^z\rvert \leqslant 1 on the closed left half-plane, and z^5 takes purely imaginary values on the imaginary axis. That makes it a straighforward application of Rouché's theorem too. For problem (iii), note that f is real (that is, f(\mathbb{R}) \subset \mathbb{R}, hence f(\overline{z}) = \overline{f(z)} ... 2 Since r is a root of f and -r is a root of g,$$\begin{aligned} 0 &= r^2 + 17r + a\\ 0 &= (-r)^2 - 17(-r) - a = r^2 + 17r - a \end{aligned}$$Combining both equations,$$\begin{aligned} 0 &= 2r^2 + 34r \end{aligned}$$So$$\begin{aligned} 0 &= 2r(r + 17) \end{aligned}$$which shows that r = 0 and r = -17 are possible roots ... 1 You can also see this from the quadratic formula. For  \ f(x) \ ,$$ x_f \ = \ \frac{-17 \ \pm \sqrt{17^2 - 4a}}{2} \ , $$while for  \ g(x) \ ,$$ x_g \ = \ \frac{17 \ \pm \sqrt{17^2 + 4a}}{2} \ . $$In order to meet the stated condition, either both discriminants must be zero, which is not possible, or they must be equal, which requires  \ a \ ... 1 You can only approximate an L^p(Ω) function by a polynomial sequence if the domain Ω is a pre-compact set (if not, no polynomial will be in that space), and to be meaningful, of Lebesgue-measure greater than zero. Then the coefficients of the polynomial p can be obtained from the moments \int_Ω x^k\,p(x)\,dx of the polynomial. And since the moments ... 1 There are two possible approaches: 1) By using repeatedly the division algorithm for univariate polynomials. In this case note that \rho(a) is in fact generated by a_0T_i-a_iT_0 with i=1,\dots, n and write$$f(T_0,T_1,\dots,T_n)=(a_0T_1-a_1T_0)g(T_0,T_1,\dots,T_n)+f_1(T_0,T_2,\dots,T_n)$$and so on. 2) By using the division algorithm for multivariate ... 1 Let g(X,Y) = (X-Y-Y^{p-1})(X-Y^2-Y^{p-2})\cdots (X-Y^{\frac{p-1}{2}}-Y^{\frac{p+1}{2}}) \in \mathbb{Q}[X,Y], so that g(X,\zeta)=f(X) \in \mathbb{Q}[X]\subset\mathbb{Q}(\zeta)[X]. We can identify \mathbb{Q}(\zeta)(X) with \mathbb{Q}(X)[Y]/\Phi_p (Y), where \Phi_p (Y) = (Y^p-1)/(Y-1) is the p-th cyclotomic polynomial. Since the image of g in the ... 1 This ring you have constructed is not a field, since we have (non-zero) zero-divisors. Note that$$ (a - 1)(a^2 + a + 1) = a^3 - 1 = 0 $$The unique field (up to isomorphism) of size 27 is given by$$ F_{27} = \mathbb{Z}_3[t]/(t^3 + t^2 + t + 2) $$The ring you have constructed can be thought of as$$ R = \mathbb{Z}_3[a,b]/((a^3-1)(b^3-1)) $$1$$\sum^l_{k=1}h_kf_k=h_1f_1+h_2f_2+\ldots+h_lf_l$$Since h_k>0 and f_k<0 when g_k=0,$$h_1f_1+h_2f_2+\ldots+h_lf_l<0$$I.e.,$$\sum^l_{k=1}h_kf_k<0\mbox{ when $g_k=0$, i.e., when $\sum^l_{k=1}h_kg_k=h_1g_1+h_2g_2+\ldots h_lg_l=0\forall x\neq0$}$$So,$$\boxed{\sum^l_{k=1}h_kf_k<0\mbox{ }\mathbf{ iff }\mbox{ }\sum^l_{k=1}h_kg_k=0\forall ...

1

If the degree of $P$ is less or equal $0$ then the result is clear. Otherwise, let $P(X)=\sum_{k=0}^n a_k X^k,\; n\ge1$. $$P(P(X))-X=P(P(X))-P(X)+P(X)-X=\sum_{k=0}^n a_k\left((P(X))^k-X^k\right)+P(X)-X\\=\sum_{k=1}^n a_k\left((P(X))^k-X^k\right)+P(X)-X$$ But for $1\le k\le n$, ...

1

Hints: Did you try a little induction of the sum of the degrees? That can help quite a bit... $$\sum_{i=0}^na_ix^i\sum_{k=0}^mb_kx^k=a_0b_0+x^2\sum_{i=1}^na_ix^{i-1}\sum_{k=1}^mb_kx^{k-1}$$ But then both summands on the right are polynomials the sum of which degrees is less than the original ones'...

1

The key idea of the polynomial division algorithm is this: if the leading coefficient of the divisor $= 1$ (or is invertible), and the dividend has degree $\ge$ the divisor, then we can $\rm\color{#c00}{scale}$ the divisor so that it has the same degree and leading coef as the dividend, then subtract it from the dividend, thereby canceling the leading term ...

1

Suppose $m(x)$ is the minimal polynomial of $A$. Then for any $B$, we have $m(\phi)(B)=m(A)B=0$. Therefore, $m(x)$ also annihilates $\phi$, and so minimal polynomial of $\phi$ is divisible my $m(x)$. On the other hand, if $n(\phi)=0$ for some polynomial $n(x)$, in particular $n(\phi)(I)=0$. But $n(\phi)(I)=n(A)I=n(A)=0$, so any such $n(x)$ is divisible by ...

1

The problem with your approach is that the resultant equations obtained may contain some similar equations, which may leave a wrong impression that there can be infinitely many solutions. Instead solve it this way. Comparing the coefficients of both the polynomials we get 4 equations, since the degree of each polynomial is $3$. We now have 4 equations and 4 ...

1

The polynomial equation $x^n-1=0$ has n complex roots, one of which is definitely $1$, and, if n is even, then so is $-1$. But if you divide the polynomial with $x-1$ for odd n, that root which is $1$ just disappears. And if you were already taught complex numbers in class, then you already know by now that the complex solutions to $x^n=a$ form a regular ...

1

There is a way more concrete than the proof from Atiyah-MacDonald, that lends further insight. Namely one can represent algebraic integers as eigenvalues of square-integer matrices. Such matrices have monic integral characteristic polynomials, so their eigenvalues are algebraic integers. Conversely, $\alpha\,$ is an eigenvector of its companion matrix, i.e. ...

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