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
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Why does $(1+x)^np(x)$ has at least $n+1$ terms?

$p(x)$ is a polynomial. Assume $p(x)\ne0$. How to prove that $(1+x)^np(x)$ has at least $n+1$ terms?
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Show that $D$ is surjective, but not injective.

Question: Let $P$ denote the set of polynomials $p(x) = \sum_{i=0}^n a_{i}x^i$ for $n=0, 1, 2, ....$ and real numbers $a_{i}$. Let $D:P \to P$ be the differential operator defined by $(Dp)(x) = \sum_{i=1}^n ia_{i}x^{i-1}$ and let $I:P \to P$ be the…
Joey
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GCD of a bivariate polynomial and its partial derivative.

I am stuck in the following question: $f(x, y)$ is a bivariate polynomial with coefficients in $\mathbb Z$. We have to show that $\deg(\gcd(f, f_y)) > 0$ iff $\deg(\gcd(f, f_x)) > 0$. (Here $f_x$ denotes the partial derivative with respect to…
babu123
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Unable to find the f(x)

Find the Cubic in $x$ which vanishes when $x=1$ and $x=-2$ and has values $4$ and $8$ when $x=-1$ and $x=2$ resprectively. I have proceeded like $P(x)=(x-1)(x+2)f(x)$ but I unable to find $f(x)$ to satisfy for $x=-1,2$.
gaufler
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$5x^{3}-5(p+1)x^{2}+(71p-1)x-(66p-1)=0$ has three solutions which are natural numbers?

How find all values of parameter $p \in R$ such that equation $~ 5x^{3}-5(p+1)x^{2}+(71p-1)x-(66p-1)=0$ has three solutions which are natural numbers? My try $(x - 1)(5x^2 - 5px + 66p - 1)=0$ and what next?
piteer
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Prove $f(x)$ has no rational number roots

Let $f(x)=ax^2+bx+c, \ a,b,c\in \mathbb{Z}$, and such $$|f(1)|,|f(2)|,|f(3)|,|f(4)|,|f(5)|$$ are prime numbers, show that $f(x)=0$ has no rational number roots
user223800
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Proof that bernstein-coefficients of $p(x)=x$ are $b_i=a+i\frac{b-a}{n},\ i=0,...,n$

I want to proof that the bernstein-coefficients for $p(x)=x$ on $[a,b]$ are described by $$b_i=a+i\frac{b-a}{n},\ i=0,...,n$$ Where the Bernstein polynomials on $[a,b]$ are defined by $$B_i^n(x;a,b)=(b-a)^{-n}{n\choose i}(b-x)^{n-i}(x-a)^i$$ I…
rtur
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Interpolation of polynomial.

Suppose we have the polynomial $f(x)=x^3$. We can now interpolate it using the values: $$f(1)=1,f(2)=8,f(3)=27,...$$ Using only one value, we get a constant: $$f_1(x)=1,\;\{1,1,1,...\}$$ Now using two values, we can get a…
RE60K
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Proving a property of two polynomials when one of them divides the another

Suppose $P(x)$ is a polynomial which can be factored into a product of different linear terms, that is $P(x)=(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_k)$ and suppose $Q(x)$ divides $P(x)$ (i.e. $Q \mid P$). How one can prove that $Q(x)$ can also be…
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If $r$ is integer solution of $(x-a_1)(x-a_2)\cdots\cdots(x-a_{2n})+(-1)^{n-1}(n!)^2=0$ then $2nr=a_1+ a_2+\cdots+a_{2n}.$

Suppose that $a_1, a_2,\ldots, a_{2n}$ are distinct integers such that $$(x-a_1)(x-a_2)(x-a_3)\cdots\cdots(x-a_{2n})+(-1)^{n-1}(n!)^2=0$$ has an integer root $r$. Show that $$2nr=a_1+ a_2+\cdots+a_{2n}.$$
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Solutions of sixth order polynomial equations

Do you know a way to solve exactly a general sixth order polynomial equation: $x^{6}+a_{5}x^{5}+a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}=0$ ? According to this link, it is possible to solve it in terms of Kampé de Fériet functions, but in the…
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Partial Fraction Decomposition of a Polynomial division

Question :Write $$\frac{x^5}{(x^2+1 )(x+1)^2}$$ as a sum of partial fraction What I've tried is to do polynomial long division twice to reduce the degree of numerator to be smaller than denominator than carry on to the normal steps of partial…
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Finding value of a polynomial

If it is given that f(x) is a 5th degree polynomial such that $\ f(1)=1;f(2)=3;f(3)=5;f(4)=7;f(5)=9$ what is the value of f(6)? I usually solve these kind of problems by making equations but no luck and also is there a smarter way to do this beside…
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what is the decomposition of $(x^2+y^2+z^2)(x+y+z)(x+y-z)(-x+y+z)(x-y+z)-8x^2y^2z^2$?

I want to have a decomposition of this : $$(x^2+y^2+z^2)(x+y+z)(x+y-z)(-x+y+z)(x-y+z)-8x^2y^2z^2$$ I have tried all possible calculation which came to my mind,I will describe one of it which is better but no result: I put $(x+y-z)=a$ ,$ (-x+y+z)=b$…
kpax
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Compute the largest root of $x^4-x^3-5x^2+2x+6$

I want to calculate the largest root of $p(x)=x^4-x^3-5x^2+2x+6$. I note that $p(2) = -6$ and $p(3)=21$. So we must have a zero between two and three. Then I can go on calculating $p(\tfrac52)$ and see that the zero must lie in the interval…
user30523
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