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30

Divide $x^4-4x^3-x^2-8x+4=0$ with $x^2$ in order to get the following $x^2-4x-1-\frac{8}{x}+\frac{4}{x^2}=(x+\frac{2}{x})^2-4(x+\frac{2}{x})-5=(x+\frac{2}{x}-5)(x+\frac{2}{x}+1)$. Finally, result is $(x^2-5x+2)(x^2+x+2)$.


7

Using the equality $\cos(\pi \pm x)=-\cos(x)$ you get $$x_1=-\frac{\cos \frac{3 }{22} \pi}{\cos \frac{1}{22} \pi} \\ x_2=-\frac{\cos \frac{9}{22} \pi}{\cos \frac{3}{22} \pi} \\ x_3=-\frac{\cos \frac{15}{22} \pi}{\cos \frac{5}{22} \pi} \\ x_4=-\frac{\cos \frac{21}{22} \pi}{\cos \frac{7}{22} \pi} \\ x_5=-\frac{\cos \frac{27}{22} \pi}{\cos \frac{9}{22} \pi} $$ ...


6

We prove the following slightly more general statement: For $a,b,c,d,e\in\mathbb Q$, if $x=ap^{1/5}+bp^{2/5}+cp^{3/5}+dp^{4/5}+e$, then $x$ has a minimal polynomial of degree at most $5$. (By moving powers of $p$ into/out of the coefficient, we can restrict the exponents $k,l,m,n$ to the set $\{0,1,2,3,4\}$.) Consider the vector space $V$ with basis $\...


6

Our cubic is irreducible over the rationals, so $\mathbb{Q}(\alpha)$ has degree $3$ over the rationals. If $i$ were in $\mathbb{Q}(\alpha)$, then $\mathbb{Q}(\alpha)$ would have a subfield of degree $2$ over the rationals. This is impossible, since $2$ does not divide $3$.


6

Since $|\sin x|\leqslant1$, we should just look at regions in which $\tfrac x2\in[-1,1]$, i.e. when $x\in[-2,2]$. There's the obvious solution of $x=0$, and $x=\pm2$ yields nothing, so we'll split up the interval into two pieces $(-2,0)$ and $(0,2)$. We'll consider the function $h:x\mapsto 2\sin x-x$. Its derivative $h'$ has in these two different intervals ...


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 $x^5 + 2 x + 1$ can't be written in terms of radicals. See e.g. Abel-Ruffini theorem


5

If $a\in K^\times$ is not a quadratic residue, then the bijection $\mu_a:K^\times\to K^\times$ given by multiplication by $a$ takes any quadratic residue to a non-residue, and therefore any non-residue to a residue (since exactly half of $K^\times$ are quadratic residues; this may be proven using cyclicity). Now note that $\mu_2(3)=6$. Reformulation using ...


5

There can be as many roots in this interval as you like. Namely, start with the polynomial $f_1(x) = -x^2-x+1$ which has one root $\phi^{-1}$ strictly between $0$ and $1$. Next we can consider $$f_2(x) = f_1(x)f_1(x^3) = x^8+x^7-x^6+x^5+x^4-x^3-x^2-x+1$$ which has all its coefficients $\pm 1$ and whose roots include $\phi^{-1}$ and $\phi^{-1/3}$. And in ...


5

$2x^4+x^3-11x^2+x+2=0$ Note that the coefficients: $2,1,-11,1,2$ are symmetrical. $2(x^4+1)+(x^3+x)-11x^2=0$ $2(x^4+4x^3+6x^2+4x+1)-7(x^3+x)-23x^2=0$ $2(x^4+4x^3+6x^2+4x+1)-7(x^3+2x^2+x)-9x^2=0$ $2(x+1)^4-7(x+1)^2x-9x^2=0$ $2\left(\frac{(x+1)^2}{x}\right)^2-7\left(\frac{(x+1)^2}{x}\right)-9=0$


5

$$\sqrt{x+2}+x=0$$ $$\sqrt{x+2}=-x$$ But because $\sqrt{x}$ is taken as the principal root we need $-2 \leq x \leq 0$. That is the difference. Squaring introduces extraneous solutions.


5

One approach is to represent the roots in polar form, and take advantage of their symmetry. For the sake of ease of exposition, let the quadratic be monic; that is, $a = 1$. If it's not already in that form, it's trivial to convert it to a monic form. If the roots are real—if $b^2-4c \geq 0$—I assume you know how to handle that. So we'll just consider ...


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.


4

If $\alpha,\beta$ are the roots of $x^2-6x+7=0$, then all quadratic equations with roots $\alpha+\frac{1}{\beta}$ and $\beta+\frac{1}{\alpha}$ are $$a\left(x-\left(\alpha+\frac{1}{\beta}\right)\right)\left(x-\left(\beta+\frac{1}{\alpha}\right)\right)=0,$$ where $a\in\mathbb R$, $a\neq 0$ (saying "the equation" is wrong, because there are infinitely many of ...


4

As $\alpha\beta=\dfrac71,$ let $y=\dfrac{\alpha\beta+1}\alpha=\dfrac{7+1}\alpha\iff\alpha=\dfrac8y$ Put the value of $\alpha$ in $$x^2-6x+7=0$$ and rearrange.


4

The other two real roots, quite expectedly, are $\tan \left({\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$ and $\tan \left({2\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$. The complex roots are produced in a similar manner, ...


4

We may take $\xi$ as a primitive $19$-th root of unity, $\xi=\exp\left(\frac{2\pi i}{19}\right)$, then check that the elementary symmetric functions of $$ a = \xi^2+\xi^3+\xi^5+\xi^{14}+\xi^{16}+\xi^{17} $$ $$ b = \xi^4+\xi^6+\xi^9+\xi^{10}+\xi^{13}+\xi^{15} $$ $$ c = \xi^1 + \xi^7 + \xi^8+\xi^{11}+\xi^{12}+\xi^{18} $$ simplify giving the coefficients of $x^...


4

$$2x^4+x^3-11x^2+x+2 = (x-2)(2x^3+5x^2-x-1)=(x-2)(2x-1)(x^2+3x+1)$$ Now you can solve it easily EDIT: You can use the rational root theorem to get the possible rational roots


4

An actual analytical solution will not happen here. An analytical approximation can be given as follows. For large enough negative $a$, there are two roots, each reciprocals of each other. One is large and positive, the other is small and positive. The large and positive one can be approximated: $$x=\ln(-a-e^{1/x})=\ln(-a)+\ln(1+e^{1/x}/a) \approx \ln(-a)+...


4

$ x^4 - 4x^3 - x^2 -8 x + 4$ to try to factor it as a product of two polynomials with degree two I will try this $ x^4 -4x^3 -x^2-8x+4=(x^2+ax+c)(x^2+dx+e) $ but the constant term is 4 so we have two choices $ c=1, e=4$ or $ c=2, e=2$ if you choose the second you get the equations $ a+d=-4, 4+ad=-1, 2a+2d=-8$ if you solve them you come up with a solution ...


4

The sum of the roots of the cubic is $4-i$. Alternately, the product of the roots of the cubic is $-6(1+i)$. You know two of the roots, so you know the third.


4

It does exist, it's just $0$. Your factorization is almost right, but you forgot the leading coefficient: it should be $f(x) = c_n (x - a_1) \ldots (x - a_n)$ (where $a_1, \ldots, a_n$ are the first $n$ of those $n+1$ values). Your job is to show that the polynomial has this form, and that $c_n = 0$.


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

By the rational root theorem, there are $6$ possibilities to check: $\pm\frac12,\pm1,$ or $\pm2$. The easiest values to test are $\pm1$ and neither works. The next easiest value to test is $2$, which is a solution. In addition, using Kenny Lau's observation from his comment, the equation can be re-written as $$2x^2+x-11+\frac1x+\frac2{x^2}=0$$ Since $x=...


3

Since, roots are complex $b^2 \leq4c$. Hence, $c\geq0$. The roots are $$z_1,_2=(-b\pm i \sqrt{(4c-b^2)})/2$$We first assume roots lie in unit circle. Hence,$|z|^2\leq1$ which implies $$(b^2+(4c-b^2))/4\leq1\implies c\leq1, i.e, |c|\leq1$$ Also, $(1-c)^2\geq0\implies(1-c)^2+4c\geq4c\implies(1+c)^2\geq4c\implies(1+c)^2\geq b^2\implies|b|\leq 1+c$ Now, let's ...


3

$x = p(1 + (\frac qp)^{\frac 15})^5 \in \Bbb Q((\frac qp)^{\frac 15})$. 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 "...


3

In A Note on Trigonometric Algebraic Numbers by D. H. Lehmer (and also here with a different notation) we find that $$ z^{-d}\Phi_n(z) = \psi_n(z+z^{-1}) $$ where $\Phi_n$ is the $n$-th cyclotomic polynomial and $d=\frac{\phi(n)}{2}$ is the degree of $\psi_n$, which is half the degree of $\Phi_n$. Lehmer proves that $\psi_n$ is irreducible. The roots of $\...


3

By the translation $x\to x-\frac13$, you cancel the square term. $$(x-\frac13)^3+(x-\frac13)^2-6(x-\frac13)-7=x^3-\frac{19x}3-\frac{133}{27}.$$ Then multiplying by $27$ and replacing $x\to3x$, you get $$x^3-57x-133=0.$$ Next, setting $x=2\sqrt{19}\cos(\theta)$ (with the aim to let $4\cos^3(\theta)-3\cos(\theta)=\cos(3\theta)$ appear), $$152\sqrt{19}\cos^...


3

You are right in that you need $b^2-4c\geq0$. The formula also shows that $x>0$ if and only if $$-b-\sqrt{b^2-4c}>0,$$ or equivalently $-b>\sqrt{b^2-4c}$. In particular $b$ must be negative. Squaring both sides shows that $$b^2>b^2-4c,$$ which means $c$ must be positive. Together with $b^2-4c\geq0$ this also gives $$b\leq-2\sqrt{c}.$$ So in ...


3

Real roots: $b^2\ge4c$. If real roots, same sign: $\;x_1x_2=c>0$. If real with same sign, positive: $\;x_1+x_2=-b>0$. Summing it up, the conditions are $$ b<0,\enspace c>0,\enspace b^2\ge 4c.$$ The solutions can be visualised in a $(b,c)$-plane:


3

Hint. One may use $$ z^3 + (-4 + i)z^2 + (1 - 5i)z + 6(1 + i)=(z^2-5z+6)(z+i+1) $$ found in $(\text{ii})$ then one may use $$ z^2-5z+6=(z-3)(z-2). $$



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