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
2
votes
2 answers

Roots of a polynomial.

The polynomial $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$$ has the roots $$\alpha_1,\ldots,\alpha_n$$ What roots does the polynomial $$g(x)=a_nx^n+a_{n-1}bx^{n-1}+a_{n-2}b^2x^{n-2}\ldots+a_1b^{n-1}x+a_0b^n$$ has? I tried with a second degree…
2
votes
1 answer

Finding a polynomial with $f(i) = a _{i}$ $(i = 1, 2, \dots, n)$ which is monotonic increasing on $[1, n]$

Is there is a positive integer $m$, depending only on $n$, such that for any strictly increasing integer sequence $a _{1}, a _{2}, \dots, a _{n}$, there is some polynomial $f(x)$ of degree at most $m$ with rational coefficients, such that $$f(i) =…
math110
  • 93,304
2
votes
2 answers

Making sense of this formula relating to polynomials : $ P(x+n+1) = \sum_{i=0}^n (-1)^{n-1} \binom {n+1} i P(x+i)$

$$ P(x+n+1) = \sum_{i=0}^n (-1)^{n-1} \binom {n+1} i P(x+i)$$ What does the above formula do ? I came across this formula in a set of lecture notes, so the explanation following was a tad short. The following problems are to be solved using this…
2
votes
1 answer

$\sqrt[3]{2}$ satisfies $x^3-2=0$ Show that there is no polynomial $P(x)$ of degree less than 3 with $P(\sqrt[3]{2})=0$

$\sqrt[3]{2}$ satisfies $x^3-2=0$ Show that there is no polynomial $P(x)$ of degree less than $3$ with $P(\sqrt[3]{2})=0$ All coefficients are rational numbers. Is it by induction? Say, if $x$ has degree of $1$, it doesn't work; and for $x$ having…
Twilight
  • 553
2
votes
1 answer

Are there any ways of solving polynomials with no rational zeros?

I attempted to solve the equation $x^5-x+41=0$ using the rational zero theorem. I soon found out that the equation has no rational zeros. Many sources say when presented with a polynomial with no rational zeros, use the Newton approximation method.…
2
votes
1 answer

Multiplying 3 or more binomials easily

Is there a way to expand three binomials in one go; i.e. without first expanding two of them, then multiplying by the last one. so expanding: (x+2)(x+3)(x+1) without first having to expand to this: (x²+5x+6)(x+1) My aim is to find a method that…
hegash
  • 181
2
votes
2 answers

Polynomial of 4th degree

I would like to ask if someone could help me with the following equation. \begin{equation} x^4+ax^3+(a+b)x^2+2bx+b=0 \end{equation} Could you first solve in general then $a=11$ and $b=28$. I get it to this form but I…
user675788
2
votes
3 answers

How to determine the degree of a polynomial?

If $$g(x) = x^4 + x^3$$ From my understanding, the degree of the above polynomial i.e. $g(x)$ is 4. However, for this polynomial, $$f(x) = (x-1)(x-2) \cdots (x-p+1)$$ What degree does $f(x)$ have? My guess was, by plug in $p = 3$, I have $f(x) =…
roxrook
  • 12,081
2
votes
2 answers

Minimal polynomial- fields

Let $\zeta$ = $\cos(\frac{2\pi}{7} ) + i\sin(\frac{2\pi}{7} )$ , let $\alpha = \zeta +\zeta^{-1} $ note that $\zeta^{-1} =\zeta^6 $ I try to find the minimal polynomial of $\alpha$ over $\mathbb Q$. I only managed to show that the degree of the…
2
votes
3 answers

A Question About A Polynomial Problem

I'm trying to understand the solution to this problem:https://i.stack.imgur.com/HOGDV.jpg. The solution states that the answer is $C$ and $D$, because if $-p(x)=p(y)$, then the equation of the values that satisfy the $-p(x)=p(y)$ must by symmetric…
Mathguy
  • 33
2
votes
5 answers

Convert standard form to factor form

Im so confused now I have no idea how to factor a standard form i have example shown below Standard form $$x^4+6x^3-x^2-6x$$ Factor form $$x(x+6)(x+1)(x-1)$$ Can you explain a little bit I have no idea Thank you
Jerson
  • 191
2
votes
2 answers

Is there a polynomial with rational coefficients $P(x)$ such that $\frac {P(n+1)}{P(n)}\in\mathbb{N}$ for every $n\in\mathbb{N}_+$?

Does there exist a polynomial with rational coefficients $P(x)$ with the property $\frac {P(n+1)}{P(n)} \in \mathbb{N}$ for every $n \in \mathbb{N}_+$? If so how can I construct such a polynomial?
Kinheadpump
  • 1,331
2
votes
1 answer

Given roots, does there exist a polynomial with "integer" coefficients?

Given roots of form $$a, ~~\sqrt[3]{b+c\sqrt{d}}, \sqrt[3]{b-c\sqrt{d}}$$ $a,b,c,d\in {\mathbb Z}$, does there exist a cubic polynomial with integer coefficients with above roots? I tried few examples and I seem to find a cubic polynomial every…
AgentS
  • 12,195
2
votes
1 answer

Evaluating a polynomial multiplication over finite ring

I have 2 polynomials $P(x)$ and $Q(x)$ over a finite ring $R(x)=x^p +1$. Let $M(x)=P(x) Q(x)$ over that ring. Example: $P(x)=x-1$, and $Q(x)=x^2+x+1$, then $M(x)=x^3-1=-2$ over that ring $R(x)=x^3+1$ My question: Given a value $\alpha$, I can…
Niko
  • 21