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 $R[X_1,\ldots,X_n]$.

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On the problem of polynomial bijection from $\mathbb Q\times\mathbb Q$ to $\mathbb Q$

The question titled "Polynomial bijection from $\mathbb Q\times\mathbb Q$ to $\mathbb Q$" which was posed on MathOverflow attracted quite a lot of attention (and may be the question with most wrong answers ever asked on the website according to the…
Dal
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Is $ f_n=\frac{(x+1)^n-(x^n+1)}{x}$ irreducible over $\mathbf{Z}$ for arbitrary $n$?

In this document on page $3$ I found an interesting polynomial: $$f_n=\frac{(x+1)^n-(x^n+1)}{x}.$$ Question is whether this polynomial is irreducible over $\mathbf{Q}$ for arbitrary $n \geq 1$. In the document you may find a proof that…
Peđa
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Appearance of Formal Derivative in Algebra

When studying polynomials, I know it is useful to introduce the concept of a formal derivative. For example, over a field, a polynomial has no repeated roots iff it and its formal derivative are coprime. My question is, should we be surprised to see…
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Decomposing polynomials with integer coefficients

Can every quadratic with integer coefficients be written as a sum of two polynomials with integer roots? (Any constant $k \in \mathbb{Z}$, including $0$, is also allowed as a term for simplicity's sake.) (In other words, is any given $P(x) = A +…
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Is There A Polynomial That Has Infinitely Many Roots?

Is there a polynomial function $P(x)$ with real coefficients that has an infinite number of roots? What about if $P(x)$ is the null polynomial, $P(x)=0$ for all x?
Карпатський
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Why is it so hard to find the roots of polynomial equations?

The question that follows was inspired by this question: When trying to solve for the roots of a polynomial equation, the quadratic formula is much more simple than the cubic formula and the cubic formula is much more simple than the quartic…
Ami
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Does multiplying polynomials ever decrease the number of terms?

Let $p$ and $q$ be polynomials (maybe in several variables, over a field), and suppose they have $m$ and $n$ non-zero terms respectively. We can assume $m\leq n$. Can it ever happen that the product $p\cdot q$ has fewer than $m$ non-zero terms? I…
Chris Brooks
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Methods to see if a polynomial is irreducible

Given a polynomial over a field, what are the methods to see it is irreducible? Only two comes to my mind now. First is Eisenstein criterion. Another is that if a polynomial is irreducible mod p then it is irreducible. Are there any others?
user977
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If $f(x)=x^2-x-1$ and $f^n(x)=f(f(\cdots f(x)\cdots))$, find all $x$ for which $f^{3n}(x)$ converges.

Let $f:\mathbb{R}\to\mathbb{R}$ be the polynomial defined by $$f(x)=x^2-x-1$$ and let $$g_0(x)=f(x),\quad g_1(x)=f(f(x)),\quad\ldots\quad g_n(x)=f(f(f(\cdots f(x)\cdots)))$$ The positive root of $f(x)$ is the famous golden…
Simon Parker
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Does convergence of polynomials imply that of its coefficients?

Let $\{p_{n}\}$ be a sequence of polynomials and $f$ a continuous function on $[0,1]$ such that $\int\limits_{0}^{1}|p_{n}(x)-f(x)|dx\to 0$. Let $c_{n,k}$ be the coefficient of $x^{k}$ in $p_{n}(x)$. Can we conclude that $\underset{n\rightarrow…
Kavi Rama Murthy
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Irreducible polynomial which is reducible modulo every prime

How to show that $x^4+1$ is irreducible in $\mathbb Z[x]$ but it is reducible modulo every prime $p$? For example I know that $x^4+1=(x+1)^4\bmod 2$. Also $\bmod 3$ we have that $0,1,2$ are not solutions of $x^4+1=0$ then if it is reducible the…
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Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$

This is a question from Math Olympiad. If $\{x,y,z\}\subset\Bbb{R}^+$ and if $$x^2 + xy + y^2 = 3 \\ y^2 + yz + z^2 = 1 \\ x^2 + xz + z^2 = 4$$ find the value of $xy+yz+zx$. I basically do not know how to approach this question. Please let me know…
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Every polynomial with real coefficients is the sum of cubes of three polynomials

How to prove that every polynomial with real coefficients is the sum of three polynomials raised to the 3rd degree? Formally the statement is: $\forall f\in\mathbb{R}[x]\quad \exists g,h,p\in\mathbb{R}[x]\quad f=g^3+h^3+p^3$
Glinka
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why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$

why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$? I know this is well-known. But how to prove it rigorously? Even mathematical induction does not seem so straight-forward. Thanks.
Qiang Li
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Only 12 polynomials exist with given properties

Prove that there are only 12 polynomials that have all real roots, and whose coefficients are all $-1$ or $1$. (zero coefficients are not allowed, and constant polynomials do not count.) Two of them are obviously $x-1$ and $x+1$. I tried…
VividD
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