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3 votes

Interesting third degree polynomial

The problem is the same as showing that there exist integers $x$ such that $$P(x)^2 \not \equiv P(x) \pmod{10}.$$ For the time being, I propose a bad solution. I believe that there exists a much ...
Yathi's user avatar
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1 vote

Solving an exotic trigonometric function or a sextic

I don’t know if this helps, but you can solve for $f$ equations (3) and (4) independently of the first two equations: You have $$2e\cos(2f+\omega)=e\cos \omega-\cos(f+\omega)$$ and $$2e\sin(2f+\omega)=...
David Quinn's user avatar
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0 votes

Randomly Generating Real-Rooted Polynomial Equations

Your first question about the roots being too close together can be solved by multiplying the root by some constant $c$ greater than one, as shown below: $$p(z)=\prod_{i=1}^{n}(z-cr_i).$$ Your second ...
BugDoctor1's user avatar
3 votes
Accepted

Finding the number of zeros of a function

We can rewrite $f$ as $$f(x) = \sum^N_{n=0} a_n x^n + \sum_{m=1}^M\frac{b_m}{x^m}=\dfrac{a_0x^M+a_1x^{M+1}+\cdots+a_Nx^{N+M}+b_1x^{M-1}+\cdots+b_M}{x^M}$$ Thus, the zeroes of $f$ will be the same as ...
Julio Puerta's user avatar
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1 vote

$\tilde f_i = f_i + g_i$, then $\dim \mathbb C[t_1,\dots, t_d]/\langle f_i \rangle \leq \dim \mathbb C[t_1,\dots,t_d]/ \langle \tilde f_i \rangle$

First, note that if the inequality \begin{equation*} \dim \mathbb{C}[t_1.\dots,t_d]/I \leq \dim \mathbb{C}[t_1,\dots,t_d]/\tilde{I} \end{equation*} would always hold, then it would have to be an ...
Hugues Bellemare's user avatar
2 votes

If $4x^{10}-x^9-3x^8+5x^7+kx^6+2x^5-x^3+kx^2+5x-5 $ when divided by $(x+1)$ gives a remainder of -14, then the value of k equals?

Let $f(x) = 4x^{10}-x^9-3x^8+5x^7+kx^6+2x^5-x^3+kx^2+5x-5$. Now, since it is given that $(x+1)$ gives a remainder $-14$ when $f(x)$ is divided by it, then $f(-1) = -14$. So, placing $-1 $ in place of $...
Auliviya Das 's user avatar
0 votes

Algebraic proof of the chain rule?

Slightly more generally as to what Matt Samuel did, one can even show the multivariate chain rule. Let $A$ be any commutative unital ring. Fix a positive integer $n$. For $1\leq r\leq n$, define $\...
Elías Guisado Villalgordo's user avatar
1 vote

Coefficient of $x^{21}$ in $(1+x+x^2+\dots+x^{10})^4$

Alternative approach Consider the following tableau ...
user2661923's user avatar
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3 votes
Accepted

If $P(x)$ is a fifth degree polynomial such that $P(x)+1$ is dividable by $(x-1)^3$ and $P(x)-1$ is dividable by $(x+1)^3$, How to find $P(x)$

We shouldn't typically expect solutions for systems like the one you wrote, which has $5$ equations but only $4$ unknowns. On the other hand, your ansatzes are not full general polynomials satisfying ...
Travis Willse's user avatar
0 votes

Are there any problems about the difference between set theoretic definitions of polynomials?

Let's just look at the case of a field. There are, as you say, two possible ways to define polynomials over a field $K$: you can define a polynomial $f : K \to K$ to be a formal expression: $f(x) = ...
Rob Arthan's user avatar
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1 vote

How to factor $(x-y)^5 + (y-z)^5 + (z-x)^5$

Let $u = x - y$ and $v = y - z$. Then $u + v = x - z$, so $z - x = -(u + v)$, and: $$(x-y)^5 + (y-z)^5 + (z-x)^5$$ $$=u^5 + v^5 - (u + v)^5$$ $$=u^5 + v^5 - (u^5 + 5u^4v + 10u^3v^2 + 10u^2v^3 + 5uv^4 ...
Dan's user avatar
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0 votes

$f$ is irreducible if the polynomial reduced $p$ is irreducible and the degrees are the same

This is a standard argument. I assume you mean $f\in\Bbb{Z}[x]$, otherwise it makes no sense to talk about reducing modulo $p$. Let us proceed by contraposition: If $f$ is reducible in $\Bbb{Q}[x]$, ...
SomeCallMeTim's user avatar
2 votes
Accepted

Probability of choosing a constant polynomial

Let us use only whole numbers between $0$ & $(m-1)$ : Each Co-efficient has $m$ choices. Total number of Polynomials : $m^n$ Total number of Constant Polynomials : $m$ Probability : $P=1/m^{n-1}$ ...
Prem's user avatar
  • 9,960
1 vote

How to factor $(x-y)^5 + (y-z)^5 + (z-x)^5$

Alternate solution/idea that should be helpful in the future. Let $a,b,c$ such that $a+b+c = 0.$ Then, they are the roots of the cubic: $$P(x) = (x-a)(x-b)(x-c) = x^3+x(ab+bc+ca)-abc = x^3+qx-r.$$ ...
dezdichado's user avatar
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4 votes
Accepted

How to factor $(x-y)^5 + (y-z)^5 + (z-x)^5$

OP arguments (from the question and the comments) are valid and lead to a quick result. First of all, considering the given polynomial $$ P(x,y,z)=(x-y)^5 + (y-z)^5 + (z-x)^5 $$ as a polynomial in $x$...
dan_fulea's user avatar
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5 votes

How to factor $(x-y)^5 + (y-z)^5 + (z-x)^5$

Hint: I think the problem is $(x-y)^5+(y-z)^5+(z-x)^5$ Let $a=z-y, b=z-x\implies a-b=?$ $$(a-b)^5-(a^5-b^5)=-(5a^4b-10a^3b^2+10a^2b^3-5ab^4)=-5ab(a^3-2a^2b+2ab^2-b^3)$$ Now $a^3-2a^2b+2ab^2-b^3=(a-b)(...
lab bhattacharjee's user avatar
2 votes
Accepted

Find the difference of $4b^3 + 6b - 7$ and $-12b^2 + 11b + 5$.

The difference is the quantity $4b^3+6b+7-(-12b^2+11b+5)$, so the answer is $4b^3+12b^2-5b+2$.
Dowdow's user avatar
  • 91
2 votes

Coefficient of $x^{21}$ in $(1+x+x^2+\dots+x^{10})^4$

Restatement of the Question to Make Reference Easier Find the coefficient of $x^{21}$ in $(1+x+x^2+\dots+x^{10})^4$ Solution Steps for the Specific Coefficient and Also Full Convolution For the ...
Stephen Elliott's user avatar
9 votes
Accepted

About nonnegative polynomials

This is a simple linear programming problem, which has no solutions. Here are the calculations, in Mathematica:
Iosif Pinelis's user avatar
2 votes

Deriative of algebraic implicit function is algebraic

Yes, this is true. Viewing $y$ as a function of $x$, differentiate $P(x,y)$ with regards to $x$ to get an equation of the form $A(x,y)+B(x,y)y'=0$, so that $y'=-\frac{A(x,y)}{B(x,y)}$ inside the field ...
KReiser's user avatar
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1 vote

Coefficient of $x^{21}$ in $(1+x+x^2+\dots+x^{10})^4$

$\displaystyle (1+x+...+x^{10})^4 = \sum_{0\le a,b,c,d\le 10} x^{a+b+c+d}=\sum_{k=0}^{40}\sum_{0\le a,b,c,d\le 10\\ a+b+c+d=k}n_{4}(k)x^k$ Where $n_m(k)$ is the number of ways $k$ can be written as a ...
Ayoub's user avatar
  • 1,656
6 votes

Coefficient of $x^{21}$ in $(1+x+x^2+\dots+x^{10})^4$

$$f(x)=\bigg(\frac{x^{11}-1}{x-1}\bigg)^4=\bigg(\frac{1-x^{11}}{1-x}\bigg)^4=(1-x^{11})^4(1-x)^{-4}$$ Now by binomial expansion for negative powers, we know $$(1+y)^n=\sum_{k=0}^{∞}\binom{n}{k} y^k$$ ...
Gwen's user avatar
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1 vote
Accepted

How to make polynomials with so many exponents and variables within said terms easier to solve

I have read your syllabus. Regarding the doubt in your replies, first multiply all the (-) and (+) with the terms inside the brackets to get a fully expanded polynomial. $$(-3w^5-2w+7)-(7w^4-10w^3+w+4)...
Gwen's user avatar
  • 1,282
0 votes

IF $a+b+c=0$ prove that $ \frac{a^5 +b^5 +c^5}{5}=\frac{a^3 + b^3 +c^3}{3}\frac{a^2 + b^2 +c^2}{2}=abc\frac{a^2 +b^2 +c^2}{2} $

There are exactly two identities in this form: if $a + b + c = 0$, then $$\frac{a^2 + b^2 + c^2}{2}\cdot \frac{a^3 + b^3 + c^3}{3} = \frac{a^5 + b^5 + c^5}{5},$$ and $$\frac{a^2 + b^2 + c^2}{2}\cdot \...
River Li's user avatar
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5 votes
Accepted

How to factor $k^2(k+1)^2 -c(k+1)^2 +k^2c +c^2 +c$

There is a typo. We have to deal with the (corrected) expression $$k^2(k+1)^2-c(k+1)^2\color{red}-k^2c+c^2+c$$ Expanding the first term doesn't help much, but expanding the second term does give a ...
ultralegend5385's user avatar
2 votes
Accepted

What's the maximum and minimum value of $P(-2)$?

Your reasoning is correct, the maximum is $41$. $$P(-2) = 16a - 8b + 4c - 2d + e$$ What we need to do to find the maximum is try to make $a,c,e$ as big as possible and $b,d$ as small as possible. So $...
NtLake's user avatar
  • 175
1 vote

Upper bound of depressed quartic

The following result has not been officially recognized by the mathematical community, and all I can do is guarantee that it is absolutely correct. If the depressed quartic $x^4+ax^2+bx+c=0$ has four ...
George Plousos's user avatar
1 vote

How to solve the roots of the following function numerically?

Use SymPy: ...
Parcly Taxel's user avatar
0 votes

Find all integers such that $2n^2+1$ divides $n^3+9n-17$

If $2n^2 +1$ divides $n^3 + 9n -17$, then it also divides $2\cdot \left(n^3 + 9n -17\right)$, and thus, it also divides $2\cdot \left(n^3 + 9n -17\right) - n \cdot \left(2n^2 + 1\right)$ which is ...
Shubhav Jain's user avatar
2 votes

IF $a+b+c=0$ prove that $ \frac{a^5 +b^5 +c^5}{5}=\frac{a^3 + b^3 +c^3}{3}\frac{a^2 + b^2 +c^2}{2}=abc\frac{a^2 +b^2 +c^2}{2} $

emphasis on Vieta formulas and recurrence relation.... Let $\sigma_2 = bc + ca + ab$ and $\sigma_3 = abc. $ We are told $\sigma_1 = a+b+c = 0$ Then $a,b,c$ are the three roots of $$ x^3 + \...
Will Jagy's user avatar
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2 votes
Accepted

IF $a+b+c=0$ prove that $ \frac{a^5 +b^5 +c^5}{5}=\frac{a^3 + b^3 +c^3}{3}\frac{a^2 + b^2 +c^2}{2}=abc\frac{a^2 +b^2 +c^2}{2} $

All these operations have been done in accordance to the fact that $a+b+c=0$. $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=-2(ab+bc+ca)..(i)$$ $$a^3+b^3+c^3=3abc...(ii)$$ Then $$(i)×(ii)=(a^2+b^2+c^2)(a^3+b^3+...
Gwen's user avatar
  • 1,282
1 vote
Accepted

Irreducible polynomials with complex root.

Let $h(x)=\gcd(f,g)$. Since $f$ is irreducible, either $h$ is a unit, or $h$ is an associate of $f$. But it cannot be a unit: if $\gcd(f,g)=1$, then there exist polynomials $p(x)$ and $q(x)$ in $\...
Arturo Magidin's user avatar
2 votes
Accepted

Real coefficients polynomial $P_n(\alpha)=0,|\alpha|=1\implies \alpha^{n+1}=1$

Note that $(1-z)P(z)=a_0+(a_1-a_0)z+\cdots+(a_n-a_{n-1})z^n-a_nz^{n+1}$, so $$0=|(1-\alpha)P(\alpha)| \ge a_0-|(a_1-a_0)\alpha+\cdots+(a_n-a_{n-1})\alpha^n-a_n\alpha^{n+1}|.$$ But $|(a_1-a_0)\alpha+\...
Conrad's user avatar
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1 vote
Accepted

If $f\in \mathbb{Q}[x,y]$ factors in some field extension of $\mathbb{Q}$, then it also factors in some finite extension of $\mathbb{Q}$?

I think it is true. For every field extension $E$ of $\mathbb{Q}$, we still have that $E[x,y]$ is an UFD. Thus there are only finitely many monic factors of $f\in\mathbb{Q}[x,y]$ in $E[x,y]$. Thus by ...
William Turner's user avatar
1 vote
Accepted

Sequence $f(k)=\underbrace{\sqrt{m+\sqrt{m+\sqrt{m+\cdots}}}}_{\text{$k\ m$'s}}-\underbrace{\sqrt{m-\sqrt{m-\sqrt{m-\cdots}}}}_{\text{$k\ m$'s}}$

Consider $A(k,m)=A(k), k \ge 1$ the first term with all pluses where we omit the dependence on $m$ when fixed and $B(k,m)=B(k), k \ge 1$ the second term with the alternating minuses. We clearly have $$...
Conrad's user avatar
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1 vote
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Factorization and irreducibilty for $x^n-2x^m+1$ trinomials.

This is a consequence of known results on irreducibility and factorization of trinomials, e.g. see Schinzel's theorem below, excerpted from this 2020 survey by Koley and Reddy. Beware there is typo ...
Bill Dubuque's user avatar
1 vote

Is there a unique irreducible polynomial relating two algebraic numbers beyond their minimal polynomials?

This is more of an extended comment. I'm going to work throughout in $\mathbb{Q}[x,y]$ rather than $\mathbb{Z}[x,y]$. First, I think you are putting the wrong condition on $f$. You should not merely ...
Arturo Magidin's user avatar
3 votes
Accepted

How is it possible that polynomial $x^2 +x +2$ can't be written as binomial square but can give number that is square of some number?

We distinguish between the factorization of a polynomial $P(x) \in \mathbb{Z}[x]$ in the polynomial ring, i.e. expressing it as a product of polynomials, vs. the integer factorization of $P(x_0) \in \...
Fatih Kaleoglu's user avatar
0 votes

Necessary and sufficient conditions that a cubic equation has three positive real roots

We will prove that if (2),(3) and (4) hold, the three roots must be all positive. For $x⩽0$, (2),(3),(4) imply that $x^3⩽0,px^2⩽0,qx⩽0,r<0$, so $x^3+px^2+qx+r<0$, so $x^3+px^2+qx+r$ have no ...
hbghlyj's user avatar
  • 2,454
3 votes

Prove that $Q(x)=Q(x-1)$ must be constant

If $Q(x) = Q(x-1)$ for all $x$, then the polynomial $A(x) = Q(x) - Q(a)$ for some fixed $a\in\mathbb R$ will have infinitely many roots ($x=a-n$, with $n\in\mathbb N$ is always a root). However, the ...
NtLake's user avatar
  • 337
3 votes

Prove that $Q(x)=Q(x-1)$ must be constant

In general, if a polynomial with coefficients in $\mathbb{R}$ satisfies $Q(x)=Q(x-1)$ for all $x$, then $Q(x)-Q(0)$ will have infinitely many roots. But a nonzero polynomial over a field cannot have ...
William Turner's user avatar
5 votes

Weird solution for functional equation

(a) IN SOLVING PROBLEMS LIKE THIS, MOST GENERALLY: I think they said "obviously" from the perspective of someone who already has polynomials in mind. Polynomials are common enough to the ...
Toby Saunders-A'Court's user avatar
2 votes
Accepted

A question about Dummit & Foote's explanation on the resolvent cubic and the Galois group

Consider the following. Let $\sigma \in S_4$ act on the roots $\alpha_1, \ldots,\alpha_4$. Then, we can also think about how $\sigma$ acts on the $\theta_i$'s: $$\sigma(\theta_1) = [\sigma(\alpha_1) + ...
blomp's user avatar
  • 529
0 votes

Galois Group of $\sqrt{2+\sqrt{2}}$ over $\mathbb{Q}$

This is not a new answer, but an addition to Tim Ratigan's answer. If you solve the system of equations  $ax_{1}^3$  + $bx_{1}^2$ + $cx_{1}$  +  $d$ = $x_{2}$  $ax_{2}^3$  + $bx_{2}^2$ + $cx_{2}$  +  ...
Gerard's user avatar
  • 313
1 vote
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The projection of the polynomial onto U parallel to V

Since $V$ is one dimensional, you can write $f(x)=g(x)+a(x^3+x^2+x+1)$ where $g(x)\in U$. Evaluating at $1$ and using the fact that $g(1)=0$ by definition, we have $f(1)=4a$, so $a=f(1)/4$. So to ...
Aaron's user avatar
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1 vote
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Definitions of polynomials $F(x):=f(x+c)$ and $G(x):=f(cx)$ if $c\in Z(R)$

Essentially, to have a nice polynomial ring, you want your indeterminates to commute with the coefficients. If this is not true, you get a dichotomy between polynomials as elements of a ring and their ...
SomeCallMeTim's user avatar
0 votes

Why is the Discriminant always an integer?

As the earlier answers and the counter example by @jyrki-lahtonen tells that discriminant according to your definition need not be an integer always, if your polynomial is not monic. But one can ...
Kalas678's user avatar
  • 443
2 votes

When are two polynomials with similar roots identical?

The statement is false, unless it takes into account the multiplicities of the roots as well. A simple counterexample is given by @ThomasAndrews in the comments (consider $x^2(x-1)$ and $x(x-1)^2$). ...
ultralegend5385's user avatar
0 votes

First n-element Sum of the First Order Polynomial Series

Whenever I see a progression where the change itself changes (think two degrees of change), I think quadratic. Quadratics only need 3 points to be unique. Let $f(n):=a_n$ for notation purposes. We ...
Lucien Jaccon's user avatar
0 votes

Prove that if $f=\frac{1}{(x-a_1)*(x-a_2)...}$, then $f=\frac{k_1}{x-a_1}+\frac{k_2}{x-a_2}+...$ where $k_1=\frac{1}{(a_1-a_2) ... }$ distinct $a_i$

Restatement of Question for Clarity and Mathematical Inductive Approach Towards Its Solution This answer attempts to follow John Bentin's proof that is well-suited for an advanced audience, but may no ...
Stephen Elliott's user avatar

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