= 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 \b[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.
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