 Solution

The coefficient assumptions imply a nonconstant polynomial with $p(0)<0$ and a positive leading coefficient. Thus $p(x)\to +\infty$ as $x\to +\infty$: dividing by its leading power of $x$ leaves the positive leading coefficient plus terms tending to zero. There is some $B>0$ with $p(B)>0$. A polynomial is continuous, so the intermediate value theorem supplies at least one positive root of a polynomial in $(0,B)$.

This gives neither uniqueness nor monotonicity. Multiple crossings are possible. If the degree of a polynomial is an even integer, the same leading-term argument gives $p(x)\to +\infty$ as $x\to -\infty$, so there is also a negative root of a polynomial. For an odd integer degree of a polynomial no negative root of a polynomial is forced; a linear polynomial with negative constant coefficient and positive slope is a counterexample.