How to find a polynomial from an algebraic number?

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In summary, to find a polynomial that will give 0 when a given algebraic number is substituted in, one can use the method of repeatedly applying (b - a)(b + a) = b^2 - a^2 and b^3 - a^3 = (b - a)(b^2 + ab + a^2) to eliminate the surds until only the desired number remains. This method may require making certain choices for b and a, such as choosing b = x - √2 and a = 2^(1/3) for √2+√3+√5, and b = x - √2 and a = 2^(1/3) for 2
  • #1
guysensei1
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Given some algebraic number, let's say, √2+√3+√5, or 2^(1/3)+√2, is there some way to find the polynomial that will give 0 when that number is substituted in? I know that there are methods to find the polynomial for some of the simpler numbers like √2+√3, but I have no clue where to begin for the more complex ones like the 2 I listed above.
 
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  • #2
Do you impose any restirictions on the polynomial you are looking for? The way you stated your problem P(x)=x-a where a is your number will do. Or even P(x)=0. I guess that's not what you were looking for though, so you need to make your question more precise.
Remember that every polynomial s.t.: P(a)=0 has form P(x)=(x-a)*Q(x) where Q is another polynomial. So you can start by writing it that way and thinking what Q do you need for your polynomial to do the job. My guess is that you want to restrict yourself only to polynomial with integer coefficients, which is hinted by the fact that you explicitly stated that zeroes are algebraic. In this case Viete formulas might help you.
 
  • #3
Blazejr said:
Do you impose any restirictions on the polynomial you are looking for? The way you stated your problem P(x)=x-a where a is your number will do. Or even P(x)=0. I guess that's not what you were looking for though, so you need to make your question more precise.

Gah, I should have said that all the coefficients must be integers, and P(x)=0 is not allowed. :P
 
  • #4
guysensei1 said:
Given some algebraic number, let's say, √2+√3+√5, or 2^(1/3)+√2, is there some way to find the polynomial that will give 0 when that number is substituted in? I know that there are methods to find the polynomial for some of the simpler numbers like √2+√3, but I have no clue where to begin for the more complex ones like the 2 I listed above.

Repeated use of [itex](b - a)(b + a) = b^2 - a^2[/itex] helps. Thus with [itex]b = x - \sqrt 2 - \sqrt 3[/itex] and [itex]a = \sqrt 5[/itex] we obtain [tex](x - \sqrt 2 - \sqrt 3 - \sqrt 5)(x - \sqrt 2 - \sqrt 3 + \sqrt 5) = x^2 - 2(\sqrt 2 + \sqrt 3)x + 2\sqrt 6 = p(x)[/tex] and the [itex]\sqrt 5[/itex] is gone. Now using the same trick with [itex]b = x^2 + 2\sqrt 6[/itex] and [itex]a = 2(\sqrt 2 + \sqrt 3)[/itex] (I make this choice because [itex](\sqrt 2 + \sqrt 3)^2 = 5 + 2\sqrt 6[/itex] and I already have a [itex]\sqrt 6[/itex] anyway) yields[tex]
q(x) = p(x)(x^2 + 2\sqrt 6 + 2(\sqrt 2 + \sqrt 3)x) = x^4 - 8\sqrt 6 x^2 - 20x^2 + 24[/tex] and only the [itex]\sqrt{6}[/itex] remains. Applying the trick once more yields[tex]
r(x) = q(x)(x^4 - 20x^2 + 24 + 8\sqrt{6}x^2) = (x^4 - 20x^2 + 24)^2 - 384x^4
[/tex] and all the surds are gone.

Similarly, for [itex]2^{1/2} + 2^{1/3}[/itex] one can make use of [itex]b^3 - a^3 = (b - a)(b^2 + ab + a^2)[/itex] by setting [itex]b = x - \sqrt 2[/itex] and [itex]a = 2^{1/3}[/itex]: [tex]
(x - \sqrt 2 - 2^{1/3})((x - \sqrt{2})^2 + 2^{1/3}(x - \sqrt 2) + 2^{2/3})
= (x - \sqrt{2})^3 - 2.[/tex] (Alternatively one can observe that [itex](\sqrt 2 + 2^{1/3} - \sqrt 2)^3 = 2[/itex].) Working what one should multiply [itex](x - \sqrt{2})^3 - 2[/itex] by to obtain a polynomial with integer coefficients I leave as an exercise.
 
  • #5
pasmith said:
[tex]
r(x) = q(x)(x^4 - 20x^2 + 24 + 8\sqrt{6}x^2) = (x^4 - 20x^2 + 24)^2 - 384x^4
[/tex]

This equation (when equated to 0) does not have √2+√3+√5 as a root. I think there's a mistake, but I can't find it.
 
  • #6
guysensei1 said:
This equation (when equated to 0) does not have √2+√3+√5 as a root. I think there's a mistake, but I can't find it.

It should be [itex](x^4 - 20x^2 + 24)^2 - 96x^4[/itex], on account of the fact that [itex](4\sqrt{6})^2 = 96[/itex], not [itex]384[/itex].

(Woolfram alpha confirms that [itex]\sqrt 2 + \sqrt 3 + \sqrt 5[/itex] is a root.)
 

FAQ: How to find a polynomial from an algebraic number?

What is a polynomial?

A polynomial is a mathematical expression consisting of variables and coefficients, combined using mathematical operations such as addition, subtraction, multiplication, and exponentiation. It is typically written in the form of ax^n + bx^(n-1) + ... + cx + d, where a, b, c, and d are constants and x is the variable.

What is an algebraic number?

An algebraic number is a number that is a root of a non-zero polynomial with rational coefficients. In other words, it is a solution to an algebraic equation. Examples of algebraic numbers include integers, fractions, square roots, and cube roots.

How do I find a polynomial from an algebraic number?

To find a polynomial from an algebraic number, you can use the reverse process of factoring. Start by writing out the algebraic number as a root of a polynomial equation, then use algebraic techniques such as expanding brackets and simplifying to find the polynomial that the number is a root of.

Can all algebraic numbers be expressed as polynomials?

Yes, all algebraic numbers can be expressed as polynomials. This is because by definition, an algebraic number is a root of a polynomial equation. However, some algebraic numbers may require higher degree polynomials to express them, while others may have simpler polynomial representations.

What are some methods for finding polynomials from algebraic numbers?

There are several methods for finding polynomials from algebraic numbers, including the rational root theorem, long division, and synthetic division. These methods involve manipulating the algebraic number and using algebraic techniques to find the polynomial that it is a root of.

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