# How to find a polynomial from an algebraic number?

1. Sep 19, 2014

### guysensei1

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.

2. Sep 19, 2014

### Blazejr

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. Sep 19, 2014

### guysensei1

Gah, I should have said that all the coefficients must be integers, and P(x)=0 is not allowed. :P

4. Sep 19, 2014

### pasmith

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

Similarly, for $2^{1/2} + 2^{1/3}$ one can make use of $b^3 - a^3 = (b - a)(b^2 + ab + a^2)$ by setting $b = x - \sqrt 2$ and $a = 2^{1/3}$: $$(x - \sqrt 2 - 2^{1/3})((x - \sqrt{2})^2 + 2^{1/3}(x - \sqrt 2) + 2^{2/3}) = (x - \sqrt{2})^3 - 2.$$ (Alternatively one can observe that $(\sqrt 2 + 2^{1/3} - \sqrt 2)^3 = 2$.) Working what one should multiply $(x - \sqrt{2})^3 - 2$ by to obtain a polynomial with integer coefficients I leave as an exercise.

5. Sep 19, 2014

### guysensei1

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. Sep 19, 2014

### pasmith

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

(Woolfram alpha confirms that $\sqrt 2 + \sqrt 3 + \sqrt 5$ is a root.)