Some polynomial roots question?

Question

So I was wondering that why a root that is not rational needs its conjugate for the polynomial's root in order to make a polynomial with only integer coefficient? Why can't a root, say, $1-(2)^{1/2}$ , have a root that is $2+2(2)^{1/2}$? Isn't that also possible? Any help will be appreciated!

Answer 1

Note that in the example which you give, roots $1-\sqrt{2},\,2+2\sqrt{2}$ that this would give corresponding factors $[x-(1-\sqrt{2}]$ and $[x-(2+2\sqrt{2})]$ with product

\begin{eqnarray} [x-(1-\sqrt{2}]\cdot[x-(2+2\sqrt{2})]&=&[(x-1)-\sqrt{2}]\cdot[(x-2)-2\sqrt{2}]\\ &=&(x-1)(x-2)-2\sqrt{2}(x-1)-\sqrt{2}(x-2)+4\\ &=&x^2-(3+3\sqrt{2})x+(6+4\sqrt{2}) \end{eqnarray}

Answer 2

To answer the $1-\sqrt{2}$ question: let $P(x)$ be a polynomial with rational coefficients having $1-\sqrt{2}$ as a root i.e. $P(1-\sqrt{2})=0\,$. Note that $1 \pm \sqrt{2}$ are the roots of $x^2-2x-1=0$. The Euclidian division of $P(x)$ by $x^2-2x-1$ will give a quotient $Q(x)$ and a remainder $R(x)$ with $\deg R \lt 2\,$, both with rational coefficients:

$$P(x)=(x^2-2x-1)Q(x)+r_0+r_1 x \quad\big|\quad r_0,r_1 \in \mathbb{Q}$$

Substitute $x=1-\sqrt{2}\,$:

$$0 = P(1-\sqrt{2})=0\cdot Q(1-\sqrt{2})+r_0+r_1 (1-\sqrt{2}) \;\;\implies\;\; r_0+r_1 - r_1 \sqrt{2} = 0$$

Since $\sqrt{2}$ is (known to be) irrational, the latter equality implies $r_0=r_1=0\,$. It follows that $x^2-2x-1 \mid P(x)\,$, and therefore $P(x)$ has $1+\sqrt{2}$ as a root as well.