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6

As observed in other answers, the leading coefficient of $f$, and the constant term both have to equal $1$ (if $f$ is not identically $0$). Now assume that $f$ has complex root with absolute value greater than 1. Let $z$ one of them with greatest absolute value. Now $2z^3 + z$ is also a root, by triangle inequality $$|2z^3 + z| = |z||2z^2 + 1| \geq ... 6 You can easily prove (x-1)\underbrace{(1+x+...+x^{n-1})}_S=x^n-1, which is your desired result by expanding the left paranthesis (keyword: geometric series) as follows: (x-1)S = xS-S = (x+x^2+...+x^n)-(1+x+...+x^{n-1}) = x^n-1 (telescope sum) 5 The derivative of f(x)=x^3+4x^2-1 is f\,'(x)=3x^2+8x, which is 0 at x=0 and at x=-\frac83. Since the function is cubic with a positive leading coefficient, it has a local maximum at x=-\frac83 and a local minimum at x=0. If you calculate f\left(-\frac83\right) and f(0), you should be able to tell very quickly how many real solutions the ... 5 Since \deg(2x-1)=1 then the remainder is a constant. Write$$P(x)=(2x-1)Q(x)+r$$so what's P\left(\frac12\right)? 5 Let deg(P)=n>0 and assume without loss of generality that deg(Q) \leq deg(P). Consider the polynomial R=(P-Q)P' (P' denote the derivative of P). We have :$$deg(R)\leq 2n-1 $$Now if r is a root of multiplicity k of P then r is a root of P' of multiplicity k-1 and because Q(r)=0, r is a root of P-Q. hence r is a root of R ... 4 1) If you know that every irreducible polynomial over \mathbb R has degree 1 or 2, you immediately conclude that \mathbb C is algebraically closed: Else there would exist a simple algebraic extension \mathbb C\subsetneq K=\mathbb C(a) with [K/\mathbb C]=\operatorname {deg}_\mathbb C a=d\gt 1. But then the minimal polynomial f(X)\in \mathbb ... 4 There are various ways to show that this can not be true. E.g., by Casorati-Weierstrass, the image of |z|>R under g is dense in the plane for every R>0, so the image of the same domain under f\circ g contains a dense subset of f(\mathbb{C}) which is itself dense in the plane, showing that f \circ g has an essential singularity at \infty. ... 4 The proof follows from the following Lemma: Lemma If 0<a \leq 1 \leq b then$$(2+a)(2+b) \geq 3(2+ab)$$Proof:$$(2+a)(2+b) \geq 3(2+ab) \Leftrightarrow \\ 4+2a+2b+ab \geq 6+3ab \Leftrightarrow \\ 0 \geq 2-2a-2b+2ab \Leftrightarrow \\ 0 \geq 2(a-1)(b-1) $$QED Lemma Now, lets solve the problem. Let b_i=-\alpha_1, and we can assume without loss of ... 4 Since a_j\geq 0 (1\leq j\leq n-1), all zeros are negative. Hence$$ P(x)=(x+r_1)\cdots (x+r_n)\qquad r_j>0, 1\leq j\leq n$$and r_1\cdots r_n=1. It follows from this last equality that at least one r_j is greater than or equal to 1, which gives r_1+\cdots+r_n\geq 1. It follows that$$ P(2)=(2+r_1)\cdots (2+r_n)\geq ...

3

The conclusion of the theorem for arbitrary $g$ is false for example for $R=\mathbb{Z}$, indeed for any domain that is not a field. Consider $f=X$ and $g=2X$, or generally $f=X$ and $g = aX$ where $a$ is not invertible.

3

More generally, let $a_1$, ..., $a_n$ be integers. We will show that $$f(x) = (x-a_1)^2 \cdots (x-a_n)^2 + 1$$ is not the product of two polynomials with integer coefficients (and positive degrees!). Continuing your approach, if $f(x)=p(x)q(x)$ then we have $1 = f(a_i) = p(a_i)q(a_i)$ for all $i$, so $p(a_i)=q(a_i)=+1$ or $p(a_i)=q(a_i)=-1$. Note that ...

3

Because in this case, the remained is a real number: $$f(x)=(2x-1)g(x)+r$$ Therefore $f(\frac12)=0\cdot g(\frac12)+r=r.$

3

Let $f(x)=x^7-12x^5+23x-132=g(x)(2x-1)+r(x)$ since, deg(divisor)>deg(remainder) and given deg(divisor)=1, therefore, deg(remainder)=0, or, its a real number. put $x=\frac{1}{2}$ to get $f(\frac{1}{2})=r$

3

The equation has three solutions. To prove this, look at the values of the right hand side at these values of $x: -4, -1, 0, 1$. The values alternate between positive and negative, so there is a root between each pair of $x$ values I gave you. That gives three solutions, and a cubic can have no more than three solutions.

3

Hint: $a^4+a^2+a > a^2+a > b^2+b$ if $a > b> 1$

2

We can write the equation as $$a=(b-a^2)(b+a^2+1).$$ But if $b$ is positive the right factor $b+a^2+1$ is strictly greater than $a$, so equality can never be attained.

2

The motivation is that $\sqrt{-3}$ is a root of the first and the second has roots ${-1\pm\sqrt{-3}\over 2}$, where these are understood in the more general sense of finite field elements, since the quadratic formula works for any field of characteristic not $2$. But then it is clear how to get the roots of the second from the first, we can assume $p>3$ ...

2

$\alpha \in \mathbb{Z}$ and $a,b \in \mathbb{Z}$ together with Vieta's formula for roots of quadratic equation: $$\{\alpha,\beta\} \textrm{ are roots of }x^2+ax+b+1 = 0 \implies \begin{cases}\alpha+\beta\ = -a \\ \alpha\beta\ = b+1 \end{cases}$$ Implies $\beta \in \mathbb{Z}$. So, $$a^2+b^2 = (\alpha+\beta)^2+(\alpha\beta - 1)^2 =\alpha^2\beta^2+ ... 2 A primitive polynomial root is also the minimal polynomial of a primitive root of unity in \mathbf F_7. Let \xi be a root of f. The field \mathbf F_7(\xi) is the field \mathbf F_{49} and its nonzero elements form a group of order 48. It suffices to show xi has order 48. Anyway its order can only be a divisor of 48, i.e. 1,2,4,8,16, ... 2 For the quadratic equation (1-m^2)x^2-2m^2 x-(m^2+1)=0, the product of the roots is \frac{-(m^2+1)}{1-m^2}. Since -1 is one of the roots, the other must be \frac{1+m^2}{1-m^2}. The same method can be used for any quadratic equation with one known root. Remark: Please check the calculation of the coefficients of the quadratics. For example, when ... 2 Hint \ a\!-\!b\mid P(a)\!-\!P(b) = b\!-\!c,\, and symmetrically. Look at the consequences of these divisibilities (chain them into a cycle). 2 This is Markov's inequality$$\max_{-1\le x\le 1}|p'(x)|\le n^2\max_{-1\le x\le 1}|p(x)|$$for every polynomial of degree at most n see this Markov brothers' inequality or see:http://www.sciencedirect.com/science/article/pii/0021904590901249 2 Hint: You know p(a) = q(a) and p(b) = q(b), and also from comparing leading coefficients in (x − a)^2(x − b)^2 + 1 = p(x)q(x) the leading coefficients of p,q are same. So, p(x) - q(x) is a polynomial with at least two roots (i.e., at least of degree 2, when a \neq b). The a = b case is easy to handle. 2 The monomial basis, \pmb m(x) = (1,x,x^2,x^3\ldots,x^n)^T should be easier to work with when applying the operator D^{k}_0=\frac{d^k}{dx^k}\mid_{x=0} than the Lagrange basis, \pmb l(x). Let us find the transformation between the two bases. The Lagrange interpolating polynomial P_n(x) is the sum$$ P_n(x)=\sum_{i=0}^n f_il_i(x)=\pmb f^T\pmb l(x) $$... 2 Since the question is tagged "abstract algebra" let's use a little, viz. congruences. Proof \,\ {\rm mod}\,\ x-1\!:\,\ x\equiv 1\,\color{#c00}{\overset{\rm CP}\Rightarrow}\, x^n\equiv 1^n \  thus \ x-1\mid x^n-1 where we have employed \,\rm\color{#c00}{CP} = Congruence Power Rule (or iterated Product Rule), whose simple proof is exactly the same as ... 2 Hint: Start by setting u=x^2. You'll get a cubic in u, which can always be factored. 2 Hint: Let y = x^2 for a moment and notice that y = -1 is a root to$$y^3 + 3y^2 + 4y + 2 = 0

2

For every non-zero complex number $a$ and every polynomial $P$ the roots of $P$ and $aP$ are the same. Applying this with $a=i$ answers your question.

1

I still don't really know what you are trying to say with definition 1, but I suspect you are trying to ensure some sort of uniqueness of coefficients, and apparently define the degree of a polynomial. I think you missed the mark here, so here is a simple and correct way to define a polynomial, and it's degree. Define a polynomial of degree $n\geq0$ to be ...

1

One might think the minimum degree is $1$ since given two points $(a,b)$ and $(c,d)$ with $a \neq c$ there is a line $y=mx+b$ through the two points. But the polynomial $mx+b$ has degree $1$ only provided $m \neq 0.$ Now it would depend on how the problem is interpreted. If it means what is the minimal degree which would work for all choices of two points ...

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