<s>[INST] Proof for rational roots Let $f(x) = a_0+a_1x+\cdots+a_nx^n$ be a polynomial of degree $n$ over $\Bbb Z$. [/INST] 
Part A:

Given $\frac pq$ is a rational root of a polynomial$$f(x)=a_nx^n+x_{n-1}x^{n-1}+\ldots a_0\tag1$$
Where $a_n\in\mathbb{Z}$. We wish to show that $p|a_0$ and $q|a_n$. Since $\frac pq$ is a root$$0=a_n\left(\frac pq\right)^n+\ldots+a_0\tag2$$
Multiplying by $q^n$, we have$$a_np^n+\ldots+a_0q^n=0\tag3$$
Examining this in module $p$, we have $a_0q^n\equiv 0(\text{mod} \text{ p})$. As $q$ and $p$ are relatively prime, we have $p|a_0$. And with the same logic for module $q$, we have $q|a_n$. Which completes the proof.
Note that this theorem is called the Rational Root Theorem!


Part B:
  Roots of $9x^3+18x^2-4x-8=0$

By the Rational Root Theorem, we have the possible roots as$$\begin{align*} & \pm1\pm2\pm4\pm8\\ & \pm1\pm3\\ & \implies\pm\frac 13,\pm\frac 23,\pm\frac 43,\pm\frac 83,\pm1,\pm2,\pm4,\pm8\end{align*}\tag4$$
Testing out the points, we find that $-2$ is a root. Thus, we can factor out $x+2$ from the polynomial to get$$(x+2)(9x^2-4)=0\tag5$$
For which we see the roots as $x=-2,x=\pm\frac 23$.


Part B:
  Roots of $6x^4-7x^3+8x^2-7x+2=0$

Using our handy Rational Root Theorem, we have the possible roots as$$\begin{align*} & \pm1,\pm2\\ & \pm1\pm2\pm3\pm6\\ & \implies\pm1,\pm\frac 16,\pm\frac 13,\pm\frac 12\pm2\end{align*}\tag6$$
After some handy guesswork, we see that $\frac 12$ is a root. Thus, we can factor out $2x-1$ to get$$(2x-1)(3x^3-2x^2+3x-2)=0\tag7$$
Fortunately, we can factor $3x^3-2x^2+3x-2$ by hand. We get$$3x^3-2x^2+3x-2\implies x^2(3x-2)+3x-2\implies (x^2+1)(3x-2)$$
Thus, our quartic can be factored as$$(2x-1)(3x-2)(x^2+1)=0$$
With roots $x=\frac 12,\frac 23,\pm i$
</s>