Finding polynomials with given roots

[Mashiro]

Say we have the following conditions:

For an any degree polynomial with integer coefficients, the root of the polynomial is n. There should be infinite polynomials that satisfy this condition. What is the general way to generate one of the polynomial?

 

[BvU]

Hi,

I've been taught a different definition, more like here

If the roots are $x_1, x_2, ... x_n$, then the polynomial is $\ \displaystyle \prod_{i=1}^n (x-x_i)$

$\$

 

[WWGD]

That reminds me vaguely of Hilbert's Nullstellensatz. Maybe @mathwonk or @fresh_42 can better tell you.

 

[mathwonk]

multiply any integral polynomial by (X-n)? i.e. what BvU said.

 

[WWGD]

multiply any polynomial by (X-n)? i.e. what BvU said.

But will this generate all such polynomials? How about multiplying by irreducible ( say over the Reals) ,such as $(x^2+a); a \geq 0$?

 

[mathwonk]

I have trouble understanding the question. if it means find all polynomials with n as a root, then these are exactly the polynomials divisible by (X-n), my answer. If it means (as perhaps you are implying) find all polynomials of degree n with n roots, then BvU has answered it. If it means find polys with n distinct roots, then,....... I just don't know what he/she wants.

 

[WWGD]

At any rate, I was thinking of more general terms given subsets $S$ of $\mathbb F^n$, on polynomials of several variables, to find all the polynomials that are killed in such set. Thus my reference to Nullstelensatz.

 

[fresh_42]

At any rate, I was thinking of more general terms given subsets $S$ of $\mathbb F^n$, on polynomials of several variables, to find all the polynomials that are killed in such set. Thus my reference to Nullstelensatz.

Reminds me of an old ad over here: "We cannot know what this friendly neighbor recommends, we recommend <brand name of a product against headaches, primarily based on ASA>."

We cannot know what our friendly new member - hello and @Mashiro! - might have meant, we all guessed a lot and I personally recommend the Euclidean ring property and keep Hilbert in hand if ASA will be too weak.

 

[Vanadium 50]

I am also confused by the question, but if P(x) and Q(x) are polynomials, the roots of P(x) are also roots of P(x)Q(x). There are an infinite number of such polynomials.

 

[Mashiro]

Sorry guys I think I phrase the question insanely badly. I was trying to ask, if we know some polynomial have a certain irrational root, how to find one of the polynomial with integer coefficients only.

 

[Vanadium 50]

If $a + \sqrt{b}$ is a root, then $x^2 -2ax + a^2 - b$ is a factor. Is that what you meant?

 

[Mashiro]

Agh yes that was the equation I was looking for. Thank You so much! By the way, how to start LaTeX in physics forums? I know how to write LaTeX but dont know what is the proper way to declare LaTeX in physics forums

 

[fresh_42]

Agh yes that was the equation I was looking for. Thank You so much! By the way, how to start LaTeX in physics forums? I know how to write LaTeX but dont know what is the proper way to declare LaTeX in physics forums

https://www.physicsforums.com/help/latexhelp/

It is MathJax here on the server and the main differences are the tags. It's ## for inline tags [itex] and [/itex] or the usual dollar in LaTeX, and it's $$ for the formula tags [tex] and [/tex] as it is in LaTeX.

I use additionally an AutoHotKey Script to load my keyboard with shortcuts so that I can e.g. write
\begin{align*}

\end{align*}
with the cursor placed on the empty line by hitting a single Alt+I.

 

[WWGD]

Sorry guys I think I phrase the question insanely badly. I was trying to ask, if we know some polynomial have a certain irrational root, how to find one of the polynomial with integer coefficients only.

Assuming one exists. By cardinality reasons alone, such a polynomial won't exist for every Real number. Or, specifically, for every Irrational number.

 

[mathwonk]

As WWGD says, e.g. X-π has π as root, but no integer polynomial does. As for when it does exist, the only cases I can think of, or have run into, are just when the polynomial has rational coefficients, and then of course you just multiply them out by a common denominator. But multiplying a rational polynomial by another irrational factor like (X-π) might make the coefficients of the rational factor look harder to recognize as such.
let's see (X-π)(X-1) = X^2 -(π+1)X+π. has root = 1.
or maybe X^3 +(13-3π)X^2 -(1+39π)X + 3π. has an irrational, but algebraic, root.
or X^7 - 2(π+1)X^5 + 2eX^4 +4πX^3 -4(π+e)X^2 +8πe. has root = sqrt(2).