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The Rolle theorem states that if a real function is continuous on $[u,v]$, differentiable on $(u,v)$, and has equal endpoint values, then its derivative vanishes somewhere in $(u,v)$. Order the $k$ distinct real roots as $r_1<\cdots <r_k$. The theorem gives a root of $f^{\prime }$ in each interval $(r_j,r_{j+1})$. These intervals are disjoint, so the obtained roots are distinct:

We prove the requested bound for positive roots of a polynomial by mathematical induction on the number of nonzero terms. In fact, the proof also works with roots counted according to their multiplicity. Write $V(p)$ for the number of sign changes and $N_{+}(p)$ for the positive-root count.

A one-term polynomial $a_1{x}^{d_1}$ has no positive roots and no sign changes. For the induction step, factor out the smallest power of $x$:For $x>0$ this factor is nonzero, so $N_{+}(q)=N_{+}(p)$, including multiplicities, and $V(q)=V(p)$. Differentiation removes exactly the constant term:The exponents remain strictly increasing, and the positive multipliers $e_j$ do not change coefficient signs. The induction hypothesis therefore gives

First suppose $a_1{a}_2<0$. Removing $a_1$ removes exactly one sign change, so $V(q)=V(q^{\prime })+1$. If $q$ has positive roots $r_1<\cdots <r_k$ of multiplicities $m_1,\dots ,m_k$, its derivative has multiplicity $m_j-1$ at $r_j$ and, by the Rolle theorem, an additional root between each pair. Thus the Rolle root count with multiplicities giveswhen there is a positive root. ConsequentlyIf there is no positive root, the desired bound is automatic.

Now suppose $a_1{a}_2>0$, so $V(q)=V(q^{\prime })$. Again the zero-root case is immediate. Otherwise, multiplying $q$ by $-1$ if necessary lets us assume $a_1>0$ and $a_2>0$ without changing roots or sign changes. For sufficiently small $x>0$,since $q(0)=a_1$ and the lowest-power term of $q^{\prime }$ has positive coefficient. Choose such a point $u$ before the first positive root $r_1$. The mean value theorem on $[u,r_1]$ gives a point $v\in (u,r_1)$ withBut $q^{\prime }(u)>0$. The intermediate value theorem applied to the polynomial $q^{\prime }$ therefore gives an extra root in $(u,v)$, before $r_1$. It is distinct from the roots at $r_j$ and those between consecutive roots. HenceBoth sign cases complete the induction:An initial sign change allows one extra positive root; without that sign change, the derivative must already turn before the first positive root. This is the upper-bound part of the Descartes rule of signs. The multiplicity claim used above follows by differentiating $q(x)=(x-r{)}^{m}h(x)$ with $h(r)\ne 0$: the derivative has a factor $(x-r{)}^{m-1}$ and its remaining factor is nonzero at $r$.