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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.


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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