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4

Note that the roots of $f(x)=x^2-2$ are $\pm \sqrt{2}$. Since $f(x)$ has degree $2$, if it were to be reducible then it would be the product of two linear factors, namely $f(x)=(x+\sqrt{2})(x-\sqrt{2})$. So suppose that $\sqrt{2}\in\mathbb{Q}(\sqrt{3}) \implies \exists a,b\in \mathbb{Q}: \sqrt{2}=a+b\sqrt{3}\implies 2=a^2+2ab\sqrt{3}+3b^2$ $\implies 2-a^2-...


3

If $n \neq 1,2,3,4,6$ then $\phi(n) >2$ and $$Q(x)=(x-\zeta_n)(x-\zeta_n^{n-1})$$ is a divisor of your polynomial with real coefficients. As for the edited question For each $\gcd(k,n)=1$ the polynomial $$Q(x ) = ( x-\zeta_n^k) (x-\zeta_n^{n-k}) $$ is irreducible over $\mathbb R$. Extra Here is the proof that, for $n \neq 1,2$, and all $(k,n)=1, Q(x)$ ...


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Yeah, in fact, there is a simple way to prove this result. At first, we can view the ring $\mathbb R[x,y]$ as a ring with one variable $Y$ over $\mathbb R[x]$, i.e., $\mathbb R[x, y] \cong \mathbb R[x][y]$. Note that $\mathbb R[x]$ is an integral domain and $x$ is an irreducible element in $\mathbb R[x]$. Hence, $x$ is a prime element in $\mathbb R[x]$. By $$...


3

Using Chebyshev polynomials, we can write $\cos 4\theta − \cos 3\theta =0$ as $T_4(\cos \theta) - T_3(\cos \theta)=0$. Now, $$ T_4(x)- T_3(x) = (8x^4-8x^2+1) - (4x^3-3x) = (x - 1) (8 x^3 + 4 x^2 - 4 x - 1) $$ Finally, $8 x^3 + 4 x^2 - 4 x - 1$ is irreducible over $\mathbb Q$ because it has no rational root. (Check!) Alternatively, it has no root mod $3$.


2

Observe that $f(0)=-9$ and $f(1)=7$. So by the IVT the polynomial has at least one real root. Now note that $f^\prime(x) = 45x^4+14x$. Creating a sign chart and sketching the graph, you observe that it crosses the $x$-axis exactly once. Therefore, you can conclude there is exactly one real root, and it is between $0$ and $1$. Furthermore, applying the ...


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Hint 1. A rational root $p/q$ of $f(x):=2x^3-3x^2+6$ with $\gcd(p,q)=1$ is such that $p$ divides $6$ and $q$ divides $2$ (See the Rational Root Theorem). Hint 2. Note that $f'(x)=6x(x-1)$, $x=0$ is a local maximum and $x=1$ is a local minimum with $f(1)=5>0$. Show that from this it follows that the roots are not all real.


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Take the element $x + \langle 1 + x^2 \rangle \in \Bbb{R}[x] / (x^2 + 1)$. Let $p(x) = x^2 + 1$. Then \begin{align*} p(x + \langle 1 + x^2 \rangle) &= (x + \langle 1 + x^2 \rangle) \cdot (x + \langle 1 + x^2 \rangle) + (1 + \langle 1 + x^2 \rangle) \\ &= x \cdot x + (x + \langle 1 + x^2 \rangle) + 1 + \langle 1 + x^2 \rangle \\ &= (x^2 + 1) + \...


2

If $r$ is a root of a polynomial $p(x)$ then $x-r$ is a factor of that polynomial. Conversely, if a polynomial has a linear factor then it has a root. With that information you should be able to prove that $x^2 + 1$ is irreducible over the reals. If you think a little bit you should be able to find a root of that polynomial in $\mathbb{R}[x]\space/\space(x^...


2

Hint: elementary method. If it factors, it is a product of homogeneous polynomials. Necessarily, one can obtain a factorisation as a product of a polynomial of (total) degree 2 and a linear polynomial. On the other hand, it has degree $2$ in $y$, so a factorisation, if any, can be found in the form $$y^2z-x^3=(y+\ell(x,z))(yz+q(x,z)), $$ where $\ell(x,z)$ ...


2

This is not a complete answer, but summarizes two known provable cases. For simplicity (and better familiarity with existing posts), let $f_n(x)=x^{n-1}+2x^{n-2}+\dots+(n-1)x+n$. Case 1: $n+1$ is a prime. As you have noticed, in many cases we can use Eisenstein criterion for $f_n(x+1)$. We can show that this will work when $n+1$ is a prime. We have $$ f_n(x+...


2

If $x^2-2$ is reducible, then it has a root; this means there exist $a,b\in\mathbb{Q}$ such that $(a+b\sqrt{3})^2=2$. This becomes $$ a^2+2ab\sqrt{3}+3b^2=2 $$ Since $\sqrt{3}$ has degree $2$ over $\mathbb{Q}$, we know that $1$ and $\sqrt{3}$ are linearly independent over $\mathbb{Q}$. Hence the above equality becomes \begin{cases} a^2+3b^2-2=0 \\[4px] 2ab=0 ...


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Suppose $x^n-a^n$ reducible in $F(a^n)$. Then this gives $a$ has degree $d<n$ in $F(a^n)$, i.e., $$(x-a)\mid x^d+c_{d-1}x^{d-1}+\dots+c_0,\quad c_j\in F(a^n)$$ in $F(a)$. So clearing denominators, $$p_d(a^n)a^d+\dots+p_1(a^n)a+p_0(a^n)=0\tag{*}$$ where $p_i(x)\in F[x]$. Note that $p_j(a^n)a^j$ contributes only powers of $a$ which are $j$ mod $n$ so are ...


1

Divide $f(x)$ by $p(x)$; you wll get $f(x)=p(x)q(x)+r(x)$ with $\deg r(x)<\deg p(x)$. On the other hand$$0=f(\alpha)=p(\alpha)q(\alpha)+r(\alpha)=r(\alpha).$$But $p(x)$ is a minimal polynomial and $\deg r(x)<\deg p(x)$. So, $r(x)$ can only be the null polynomial, which means that $f(x)=p(x)q(x)$. So, $p(x)\mid f(x)$. As you can see, the irreducibility ...


1

If you start with the polynomial $a=b=0$, then the polynomial will always be $x^2$ and so the process goes on forever. However, in all other cases, it will terminate in finitely many steps. First, suppose $b=0$ and $a\neq 0$. If $a<0$, the next polynomial is $x^2-a$ which does not have real roots. If $a>0$, the next polynomial is $x^2-ax$ and then ...


1

In this case, note that for $\beta=\omega+\omega^2+\omega^4$, $\beta^2= \underbrace{\omega^2+\omega^4+\omega^8}_\beta+2(\underbrace{\omega^3+\omega^5+\omega^6}_{-1-\beta})=-\beta-2$, so $\beta^2+\beta+2=0$. One other approach, which can be used more generally, if you know some Galois theory, is to calculate $Gal(\mathbb Q(\omega)/\mathbb Q)$. Then you ...


1

Well, let $I=\langle x^2+1\rangle$. The quotient ring ${\Bbb R}[x]/I$ contains a zero $\bar x =x+I$ of $x^2+1$, since $\bar x^2 + 1 = (x+I)^2+1 = (x^2+1)+I = \bar 0$ in the quotient ring. Thus the quotient ring consists of the elements $a+b\bar x$, $a,b\in{\Bbb R}$. Addition is componentwise $$(a+b\bar x) + (c+d\bar x) = (a+c) + (b+d)\bar x$$ and ...


1

The polynomial is irreducible modulo $17$. This is (a little bit) easier than to check over the integers. First, there is no root modulo $17$, and then writing the polynomial as $(x^2+ax+b)(x^2+cx+d)$ quickly gives a contradiction modulo $17$. Substituting $x$ by $x-1$ we obtain $f=x^4-3x^2+9x-11$, which is a bit easier to handle. First we have $c=-a$ and ...


1

We have $$ 2x^3-5x^2+6x-2=(x^2-2x+2)(2x-1), $$ so that this polynomial is reducible. The second one is irreducible with $p=2$ by Eisenstein, for $f(x-1)$. The transformed polynomial is $x^4-2x+2$.


1

Another way is to use the fact that $\zeta_7 = \cos \frac{2\pi}{7} + i \sin \frac{2\pi}{7}$ is a $7$-th root of unity. Now $\cos \frac{2\pi}{7} = \frac{\zeta_7 + \zeta_7^{-1}}{2}$. Now the $7$-th cyclotomic polynomial is: $$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$$ This is satisfied by $\zeta_7$. Divide by $\zeta_7^3$ to get: $$(\zeta_7^3 + \zeta_7^{-3}) + (\...


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From the comments below it appears that OP is also interested in the first half of the question (which I had originally overlooked as just a substitution). Hint For the first part of the question, we are essentially asked to show that $$\cos \left(4 \cdot \frac{2 \pi k}{7}\right) - \cos \left(3 \cdot \frac{2 \pi k}{7}\right) = 0$$ for all integers $k$. Here ...


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Well, its sufficient to check that the polynomial cannot be divided by an irreducible polynomial of degree 1, $X$ and $X+1$ (this is clear since the polynomial has no zero in the prime field), and degree 2, $X^2+X+1$ (which is the only irreducible polynomial of this degree).


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You are going to need to use numerical approximations. Newton-Raphson is good, there is also the Golden-Section search and the Bisection algorithm. If you were wondering what the answer is, using mathematicas NSolve function will use a combination of Bisection, Newton-Raphson to approximate the values. NSolve[$9x^5 + 7 x^2 - 9 == 0, x$] Which produces ...


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$$f(0)=-9\text{ and }f(1)=7$$ so that the positive root lies somewhere in between. We can use the starting value $\frac12$ for Newton's iterations, $$x\leftarrow x-\frac{9x^5+7x^2-9}{45x^4+14x}.$$


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This follows from Eisensteins's criterion, with $p=3$.


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Hint: Rational root theorem. $\hphantom{imaginary space}$


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