Find Polynomials with Integer Coefficients with Particular Roots

Question

Is there any simple way to find a polynomial with integer coefficients so that ($x=\sqrt{2} +\sqrt{3}$) is one of its roots? I know one way is to get rid of all the square roots in the equation to be left with rational numbers, but is there any other simple way?

Answer 1

Is this simple? $$x=\sqrt2+\sqrt3$$ $$x^2=2+3+2\sqrt2\sqrt3$$ $$x^2-5=2\sqrt6$$ $$(x^2-5)^2=24$$ $$x^4-10x^2+1=0$$

Answer 2

Square it $$x^2 =2+2\sqrt{6}+3$$ so $$(x^2-5)^2 = 4\cdot 6$$ so $$ p(x) = x^4 -10x^2+1$$

Answer 3

We have $x_0=\sqrt2+\sqrt3$. If we take another root to be $x_1=\sqrt2-\sqrt3$, by Vieta's formulas we get the equation

$$x^2-2\sqrt2x-1=0$$ and we can get rid of the square root by writing

$$(x^2-1)^2=8x^2.$$

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Alternatively,

$$x-\sqrt2=\sqrt3\to x^2-2\sqrt2x+2=3\to(x^2-1)^2=8x^2.$$