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
1 answer

Irreduciblity of polynomial in $\Bbb Q[x]$

I am trying to prove that $x^5 + 16$ is irreducible in $\Bbb Q[x]$. (Hint: Consider $(x-1)^5 + 16$) Not sure how to use the hint, thanks for any help in advance.
Amz
  • 127
2
votes
4 answers

What is the solution of $x^3+x=1$?

According to Wolfram|Alpha, the solution of $x^3+x=1$ is approximate $0.68233$ or exactly this monstrosity: $x_0=\frac{\sqrt[3]{\frac{1}{2}(9+\sqrt{93})}}{3^{\frac{2}{3}}}-\sqrt[3]{\frac{2}{3(9+\sqrt{93})}}$ $x^3+x=1$ is so simple, that I refuse to…
zvavybir
  • 169
  • 6
2
votes
1 answer

Solve a cubic polynomial given that one root is four times a second root?

How exactly would you solve the equation: "Solve the equation $10x^3+23x^2+5x−2=0$ given that one root is four times a second root." How would you go about solving this? Any help would be greatly appreciated.
missiledragon
  • 605
  • 9
  • 23
2
votes
1 answer

Evaluate $P(x)=5+3x+5x^2+4x^4$ using Horner's method

I want to evaluate $P(x)=5+3x+5x^2+4x^4$ using Horner's method So I rewrite $P(x)$ as : $$P(x)=5+3x+5x^2+0x^3+4x^4\tag{1}$$ and applying Horner's method to get: $$P(x)=5+x(3+x(5+x(0+4x)))\tag{2}$$ Is it done? or I should rewrite it as…
Aligator
  • 1,454
2
votes
2 answers

Why is it given that an odd polynomial can be written in the form $ax^3 +bx$?

So a question give that P(x) is an odd polynomial of degree three. So now I know that $P(x) = ax^3 + bx^2 +cx +d $. And $P(-x) = -P(x) $since it is odd. However, the solutions say that given this informations, it can actually be written simply in…
user71207
  • 1,543
2
votes
1 answer

Is it possible to find 3 quadratic polynomials $f(x),g(x),h(x)$ such that $f(g(h(x))) = 0$ has 8 roots $1,2,3,4,5,6,7,8$?

Here is what I tried - I assumed 3 quadratics $f(x) , g(x)$ and $h(x)$ with these I formed the gigantic equation $f(g(h(x)))$ it had a degree 8 therefore was satisfying the 8 degree criterion but I'm stuck at finding is it possible that it has roots…
2
votes
1 answer

Find $\sqrt{a}+\sqrt{b}+\sqrt{c}$ only in terms of $p$.

If $a^{2} x^{3}+b^{2} y^{3}+c^{2} z^{3}=p^{5}, a x^{2}=b y^{2}=c z^{2}$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{p},$ find $\sqrt{a}+\sqrt{b}+\sqrt{c}$ only in terms of $p$. I tried simple factorization like shifting terms here and there…
2
votes
2 answers

Solve a cubic polynomial?

I've been having trouble with this question: Solve the equation, $$5x^3 - 24x^2 + 9x + 54 = 0$$ given that two of its roots are equal. I've tried methods such as Vieta's formula and simultaneous equations, assuming the roots are: $a$, $a$, $b$, but…
missiledragon
  • 605
  • 9
  • 23
2
votes
1 answer

Finding the sum of the coefficients of polynomial of degree 21

Problem : Find the sum of the coefficients of the polynomial$ p(x) =(3x-2)^{17}(x+1)^4$ Solution : $ p(x) =(3x-2)^{17}(x+1)^4 $ $= (a_0+a_1x+....a_{17}x^{17})(b_0+b_1x+....b_4x^4)$ for some $a_i; b_j$ = $(c_0+c_1x+...c_{21}x^{21})$ for some $c_k$…
Sachin
  • 9,896
  • 16
  • 91
  • 182
2
votes
4 answers

Remainder Theorem: $f(x) = x^3+1$ divided by $\operatorname {d}(x)= x$

Take for example $\operatorname {f}(x) = x^3+1$ divided by $\operatorname {d}(x)= x-1$ $$\frac {x^3+1}{x-1}$$ The Remainder Theorem tells us that the remainder will be $$\operatorname {f}(zero-of-\operatorname {d}(x))$$ In this case the remainder…
HPSmash77
  • 53
  • 3
2
votes
1 answer

Parameterizing The Space of Polynomials with Linear Constraints

Suppose I am working with polynomials $p(z)$ over $\mathbb{C}$, of some degree $N$, and where $a_n=1$. I would like to find a simple way to parameterize the set for which there exists a prescribed linear condition $C(r_1,r_2,\cdots r_N)$ on the…
2
votes
1 answer

Applying the polynomial remainder theorem - does this logic hold for working out factors?

Let's say I have some function of $x$ and an unknown integer $d$, given by $f(x) = x^2 + d^2 + dx$ and I want to see if it's divisible by something like $x-d$;as I understand it, we could apply the polynomial remainder theorem and find the…
DRG
  • 367
2
votes
2 answers

Find a solution of the polynomial

Given $1<\beta<2$ and positive integer $n\geq2013$, then can we find a non-zero vector $(a_n\,,a_{n-1}\cdots\,a_1\,a_0 )$ where all $a_i\in\{-1\,,0\,,1\}$, such that $\sum\limits_{k=0}^{n-1}a_k\beta^k=0$
Tao
  • 530
  • 2
  • 11
2
votes
1 answer

$f(x)$ can't be factorize in $p(x)q(x)$ where where p and q are of degree $\le 3 $

Let $f(x)=x^4+26x^3+52x^2+78x+ 1989$ $f(x)$ can't be factorize in $p(x)q(x)$ where where $p\ and\ q$ are of degree $\le 3 $
ROBINSON
  • 2,269
2
votes
2 answers

Quadratic equation and Proof

For rational numbers $a$ and $b$, the quadratic equation $x^2 - ax - b = 0$ has two solutions according to my professor. How can I Prove that if one of these is solutions is rational, the other must be as well. We have been given a small hint: if…