Tag Info

Hot answers tagged irreducible-polynomials

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$(k,n)=1, Q(x)$... 3 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 ... 2 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. 2 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 formy^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 Ifx^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 ... 2 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: $$(\zeta_7^3 + \zeta_7^{-3}) + (\... 1 From the comments below it appears that OP is also interested in the first half of the question (which I had originally overlooked as just a substitution). Hint For the first part of the question, we are essentially asked to show that$$\cos \left(4 \cdot \frac{2 \pi k}{7}\right) - \cos \left(3 \cdot \frac{2 \pi k}{7}\right) = 0$$for all integers k. Here ... 1 Well, its sufficient to check that the polynomial cannot be divided by an irreducible polynomial of degree 1, X and X+1 (this is clear since the polynomial has no zero in the prime field), and degree 2, X^2+X+1 (which is the only irreducible polynomial of this degree). 1 You are going to need to use numerical approximations. Newton-Raphson is good, there is also the Golden-Section search and the Bisection algorithm. If you were wondering what the answer is, using mathematicas NSolve function will use a combination of Bisection, Newton-Raphson to approximate the values. NSolve[9x^5 + 7 x^2 - 9 == 0, x] Which produces ... 1$$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}.$$1 This follows from Eisensteins's criterion, with$p=3$. 1 Hint: Rational root theorem.$\hphantom{imaginary space}\$

Only top voted, non community-wiki answers of a minimum length are eligible