Questions tagged [polynomials]

For both basic and advanced questions on polynomials in any number of variables, including, but not limited to solving for roots, factoring, and checking for irreducibility.

Usually, polynomials are introduced as expressions of the form $\sum_{i=0}^dc_ix^i$ such as $15x^3 - 14x^2 + 8$. Here, the numbers are called coefficients, the $x$'s are the variables or indeterminates of the polynomial, and $d$ is known as the degree of the polynomial. In general the coefficients may be taken from any ring $R$ and any finite number of variables is allowed. The set of all polynomials in $n$ variables $X_1,\ldots,X_n$ over a ring $R$ is denoted by $R[X_1,\ldots,X_n]$. Strictly speaking this is a formal sum, because the variables do not represent any value. Nevertheless, the variables of a polynomial obey the usual arithmetic laws in a ring (like commutativity and distributivity). This makes $R[X_1,\ldots,X_n]$ a ring itself. One should note that $R[X_1][X_2]=R[X_1,X_2]$. This idea can be extended to $R[X_1,\ldots,X_n]$ in a very natural way.

An expression of the form $rX_1^{i_1}X_2^{i_2}\cdots X_n^{i_n}$ ($r\in R$) is called a term (of the polynomial). Polynomials are defined to have only finitely many terms. An expression with infinitely many different terms is generally not considered to be a polynomial, but a (formal) power series in one or more variables.

When $P\in R[X]$, $P(x)$ is the evaluation of $P$ at $x$ (pronounced $P$ of $x$, or simply $Px$). Here $x$ does not necessarily have to be an element of $R$. For $P(x)$ to be properly defined for an $x$ in some ring $S$ we need:

  • a homomorphism $\phi:R\to S$
  • the image of all coefficients of $P$ under $\phi$ should commute with $x$.

Evaluation is now simply performed by replacing all coefficients $r_i$ of $P$ by $\phi(r_i)$ and all appearances of $X$ by $x$. This quite naturally gives an expression that is well defined as an element of $S$. The concept of evaluation is naturally extended to $R[X_1,\ldots,X_n]$.

26755 questions
16
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How do we check if a polynomial is a perfect square?

Often we come across polynomial diophantine equations of the form $y^2 = x^2 + x+ 1$ and there are two ways to disprove the existence of solutions to such equations: 1) By bounding between consecutive perfect squares. 2) By reading the equations in…
Isomorphism
  • 5,693
15
votes
1 answer

Polynomials with rational zeros

Find all polynomials $F(x)={a_n}{x^n}+\cdots+{a_1}x+a_0$ satisfying $a_n \neq0$; $(a_0, a_1, a_2, \ldots ,a_n)$ is a permutation of $(0, 1, 2 ... n)$; all zeros of $F(x)$ are rational.
15
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5 answers

How to find $x^{2000}+x^{-2000}$ when $x + x^{-1} = \frac{1}{2}(1 + \sqrt{5})$

Let $x+x^{-1}=\dfrac{1+\sqrt{5}}{2}$. Find $x^{2000}+x^{-2000}$. How many nice methods do you know for solving this problem? Thank you everyone. My method: because $x+\dfrac{1}{x}=2\cos{\dfrac{2\pi}{5}}$, so…
math110
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14
votes
1 answer

Does there exist a polynomial $p(x) \in \mathbb C[x]$ such that $p(x) \notin \mathbb R[x]$ and $p(x)p(-x)=p(x^2)$?

Does there exist a polynomial $p(x) \in \mathbb C[x]$ such that $p(x) \notin \mathbb R[x]$ and $p(x)p(-x)=p(x^2)$ ? I have noticed that if $a_n$ is the leading coefficient of $p(x)$ then $a_n=(-1)^n $, so the leading coefficient cannot be…
user228169
13
votes
3 answers

Solve $3n^3 + 3n^2 + 4n = n^n$ in positive integers

So my cousin is in the math team (7th grade) and he was asking me for help on one of his problems but I don't know how to solve For what positive integer n does $3n^3 + 3n^2 + 4n = n^n$ anyone know how to do this?
13
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6 answers

Does there exist an even polynomial that has a term with an odd power?

Does there exist an even polynomial (i.e. $f(-x)=f(x)$ for all $x$) that has at least one odd power of $x$?
Superbus
  • 2,136
13
votes
1 answer

If polynomial with rational number is injective on rationals then it is injective on reals?

Let $p:\Bbb{R}\to\Bbb{R}$ is polynomial with rational coefficients. If restriction of $p$ to $\Bbb{Q}$ is injective, then $p$ is injective? I conjectured that $p$ is monotonic, but I don't know how to prove this conjecture. Thanks for any help.
Hanul Jeon
  • 27,376
13
votes
1 answer

$a,b,c,d\ne 0$ are roots (of $x$) to the equation $ x^4 + ax^3 + bx^2 + cx + d = 0 $

Find all quadruplet(s) of non-zero real numbers $ (a,b,c,d) $ such that $ a,b,c$ and $ d$ are roots (of $x$) to the equation $ x^4 + ax^3 + bx^2 + cx + d = 0 $. My friend found a set of irrational roots that satisfy this condition through…
GohP.iHan
  • 1,376
12
votes
1 answer

Bounding the "complexity" of irreducible factors of an integer polynomial

Given an integer polynomial $P(x) = a_0 + a_1 x + \cdots + a_n x^n$, there ought to be a reasonable bound on the "complexity" of its possible irreducible integer polynomial factors that allows us to restrict to finitely many possible factors. This…
Qiaochu Yuan
  • 419,620
12
votes
1 answer

Simply formulated but very hard problem about certain polynomial

Let $L:=[k_1,\dots,k_n]$ be a list of natural numbers (i.e. $\{1,2,\dots \} $) , repetitions are allowed. How to prove that the sum of the moduli of the coefficients of the polynomial $$ \prod_{j=1}^{j=n}\left(1-x^{k_j}\right)$$ is greater than or…
user64494
  • 5,811
12
votes
5 answers

Create a monic polynomial with integer coefficients with $\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{5}$ as a root.

Create a monic polynomial with integer coefficients with $\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{5}$ as a root. I want to create such polynomial but all the ways I have tried don't give a polynomial with integer coefficients and all of them give me…
Taha Akbari
  • 3,559
12
votes
2 answers

prove this $x^{m+1}|f^{(m)}(x)-x$

Let $f(x)$ is polynomial with complex coefficients,such $$x^2|f(x)-e^{\frac{2\pi i}{m}}\cdot x$$ where $m>1$ be give postive integers,and define $$f^{(1)}(x)=f(x),f^{(2)}(x)=f(f(x)),f^{(3)}(x)=f(f(f(x))),\cdots,f^{(m)}(x)=f(f^{(m-1)}(x))$$ show…
math110
  • 93,304
12
votes
5 answers

Statement about divisors of polynomials and their roots

I am a 10th grade student and there is a statement in my math book If $a$ is a root of the polynomial $f(x)$ then $(x-a)$ is a divisor of $f(x)$ Why is $(x-a)$ a divisor of $f(x)$? Can you please tell me?
12
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2 answers

An Eisenstein-like irreducibility criterion

I could use some help with proving the following irreducibility criterion. (It came up in class and got me interested.) Let p be a prime. For an integer $n = p^k n_0$, where p doesn't divide $n_0$, set: $e_p(n) = k$. Let $f(x) = a_n x^n + \cdots +…
swan
  • 163
11
votes
1 answer

The zeros of a multivariable polynomial

The other day I came across the following statement: A polynomial $f(x,y)$ of degree at most $3$ that vanishes at $8$ of the $9$ points $(x,y)$ with $x, y \in \{-1,0,1\}$ must also vanish at the $9$th point. I am wondering about how this statement…
angela o.
  • 604