The Radical conjugate Roots Theorem If the coefficients of $$p(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$$ are rational, the Conjugate Radical Roots theorem states that if the equation $p(x)=0$ has a root of the form $x=s+t\sqrt{u}$ where $\sqrt{u}$ is rational, then the equation must also have the conjugate radical, $x=s-t\sqrt{u}$, 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 $p(x)$.
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,

$\left.\begin{matrix}p_n(a-\sqrt{b})=S_n-Q_n\sqrt{b}\\S_n=Q_n=0\end{matrix}\right\}{\Rightarrow}p_n(a-\sqrt{b})=0(\therefore)$