If the coefficients of $$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ are rational, the Conjugate Radical Roots theorem states that if the equation $p(x)=0$ has a root of the form $s+t\sqrt{u}$ where $\sqrt{u}$ is irrational, then the equation must also have the conjugate radical, $s-t\sqrt{u}$, as a root.
How to prove that statement? Can anyone show it please?
Suppose that $s+t\sqrt u$ is a root of $p(x)$ and that $\sqrt u\notin\mathbb Q$. Then $s+t\sqrt u$ is also a root of the polynomial$$q(x)=x^2-2sx+s^2-t^2u,$$ since$$q(x)=\bigl(x-s-t\sqrt u)\bigl(x-s+t\sqrt u\bigr).$$By polynomial long division, you can express $p(x)$ as $q(x)q^\star(x)+r(x)$, where $q^\star(x),r(x)\in\mathbb{Q}[x]$ and the degree of $r(x)$ is smaller that $2$ (which is the degree of $q(x)$). But, since $s+t\sqrt u$ is a root of both $p(x)$ and $q(x)$, it must be a root of $r(x)$ too, and the only polynomial in $\mathbb{Q}[x]$ with degree smaller than $2$ with an irrational root is the null polynomial. Therefore, $q(x)\mid p(x)$, and, since $s-t\sqrt u$ is root of $q(x)$, this implies that $s-t\sqrt u$ is also a root of $p(x)$.
I suppose that your hypothesis is that $s,t,u$ are rational and $\sqrt{u}$ is irrationnal.
Put $Q(x)=P(s+tx)=\sum b_k x^k$. Clearly, the $b_k$ are rational, and $Q(\sqrt{u})=0$.We have $$Q(\sqrt{u})=\sum_{2|k}b_k u^{k/2}+\sqrt{u}\sum_{2\nmid k} b_k u^{(k-1)/2}=A+\sqrt{u}B$$
Now as $A,B$ are rationals and $\sqrt{u}$ not, we must have $A=B=0$, and hence $Q(-\sqrt{u})=A-\sqrt{u}B=0$, and we are done.
Will represent a polynomial with rational coefficients:
$p_n(x)=\sum_{i=0}^{n} {a_i x^i},{a_i}\epsilon\mathbb{Q},a_n \ne{0},i\epsilon\overline{1,n},n\epsilon\mathbb{N},n\ge{2}$
that has an irrational root: $p_n(a+\sqrt{b})=0, a,b\epsilon\mathbb{Q},b\gt0$
This polynomial solution becomes: $p_n(a+\sqrt{b})=a_0+a_1(a+\sqrt{b})^1+...+a_n(a+\sqrt{b})^n =0$
Analyzing the irrational factor in the k-term:
$(a+\sqrt{b})^k=s_k+q_k\sqrt{b}, s_k, q_k\epsilon\mathbb{Q}$
then a conjugated form could be deducted for the conjugate:
$(a-\sqrt{b})^k=s_k-q_k\sqrt{b}$
This result could be applied to the solution:
$p_n(a+\sqrt{b})=S_n+Q_n\sqrt{b}=0 {\Rightarrow}S_n=Q_n=0$
However,
$\left.\begin{array}{rl}p_n(a-\sqrt{b})&=&S_n-Q_n\sqrt{b}\\S_n&=&Q_n=0\end{array}\right\}{\Rightarrow}p_n(a-\sqrt{b})=0(\therefore)$
