 Solution

Put $f(X)=X^3+5X-12$. Any root in ${\mathbb{Q}}_p$ is a p-adic integer: if its valuation were negative, $X^3$ would be the unique term of least valuation.

For $p=2$, reduction modulo two has roots zero and one. The root zero is simple because $f^{\prime }(0)=5$ is odd, so it lifts uniquely. An odd integer $x$ satisfies $x^2\equiv 1(\mathrm{mod}8)$, and henceThus there is no odd $2$-adic root and the number of roots is one.

For $p=3$,All three roots are simple because $f^{\prime }(X)=3X^2+5\equiv 2(\mathrm{mod}3)$. Each lifts uniquely, giving three roots in ${\mathbb{Q}}_3$.

For $p=5$, reduction gives $X^3-2$, whose unique root is $X=3$; it is simple because $3X^2\not\equiv 0(\mathrm{mod}5)$. It lifts uniquely, so there is one root in ${\mathbb{Q}}_5$.