2. Find all prime numbers $p$ for which the polynomial

$$
q(x)=2 x^{3}-2 p x^{2}+(1-p) x+p
$$

has at least one rational root.
III/2. If $p=2$, then $q(x)=2 x^{3}-4 x^{2}-x+2=(x-2)\left(2 x^{2}-1\right)$ and has a rational root $x=2$. Now let $p$ be an odd prime. The only candidates for rational roots are $\pm 1, \pm p, \pm \frac{1}{2}$, and $\pm \frac{p}{2}$. Let's calculate all possibilities.

We have $q(1)=3-2 p$ and since $3-2 p$ is odd, $q(1) \neq 0$. Clearly, $q(-1)=-3 \neq 0$. The expression $q(-p)=-p^{2}+2 p=p(2-p)$ is also different from 0, since $p \neq 2$. Similarly, $q(p)=-4 p^{3}+p^{2}=p^{2}(1-4 p) \neq 0$ and $q\left(\frac{1}{2}\right)=\frac{3}{4} \neq 0$. We can also check $q\left(-\frac{1}{2}\right)=p-\frac{3}{4} \neq 0$ and $q\left(\frac{p}{2}\right)=\frac{-p^{3}-2 p^{2}+6 p}{4}$. Since $p$ is odd, the numerator of the last expression is odd, so the expression cannot be equal to 0. Finally, $q\left(-\frac{p}{2}\right)=\frac{-3 p^{3}+2 p^{2}+2 p}{4}$. Again, the expression cannot be equal to 0, since the numerator is odd.

For odd primes, the polynomial $q$ has no rational roots, which means the only suitable prime is $p=2$.

The conclusion that the rational roots of the polynomial can only be $\pm 1, \pm p, \pm \frac{1}{2}$, and $\pm \frac{p}{2} \ldots \ldots \mathbf{1}$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-14.jpg?height=49&width=1631&top_left_y=655&top_left_x=224)

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-14.jpg?height=57&width=1633&top_left_y=700&top_left_x=220)

Justification that the numbers $p$ and $-p$ are not roots when $p \neq 2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

Justification that the numbers $\frac{p}{2}$ and $-\frac{p}{2}$ are not roots when $p$ is odd \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots

When $p=2$, $x=2$ is a rational root of the polynomial ..............................................................

Conclusion that the only such prime is $p=2 \ldots \ldots \ldots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$