Solve in $\mathbb{Q}$ the equation $5 x^{3}+3 x^{2}+3 x-2=0$.

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Consider a possible rational solution $\frac{p}{q}$ (written in its simplest form) of the equation. We then have

$$
5\left(\frac{p}{q}\right)^{3}+3\left(\frac{p}{q}\right)^{2}+3 \frac{p}{q}-2=0
$$

so $5 p^{3}+3 p^{2} q+3 p q^{2}-2 q^{3}=0$.

We therefore have $5 p^{3}=q\left(-3 p^{2}-3 p q+2 q^{2}\right)$, so $q \mid 5 p^{3}$, and since $q$ is coprime with $p$ (and thus with $p^{3}$), we deduce by Gauss's lemma that $q \mid 5$. By a completely analogous method, we show that $p \mid 2$. The only possible rational solutions of the equation are therefore $\pm 1, \pm 2, \pm \frac{1}{5}$, and $\pm \frac{2}{5}$. We only need to check, by hand, whether each of the potential solutions is actually a solution or not, and we see that the only solution is $\frac{2}{5}$.

Remark: This method provides a way to solve any polynomial equation with integer coefficients (and thus with rational coefficients, since we can reduce to an equation with integer coefficients by multiplying by a suitable constant) in $\mathbb{Q}$.