Let $λ\_1,\ldots,λ\_m$ real numbers pairwise distinct and $μ\_1,\ldots,μ\_m$ real numbers all nonzero. We know from polynomial interpolation that for a given $r$ such that $1\leq r\leq m$, there exists an unique polynomial $R$ of degree less than $(m+r+1)$ such that $R(λ\_i)=μ\_i,1≤i≤m$ and $R'(λ\_i)=0,1≤i≤r$ .


I want to prove the existence of a real polynomial $S$ such that the polynomial
$$P(x)=R(x)+S(x)\displaystyle\prod\_{i=1}^{r}(x-\lambda\_i)^2\displaystyle\prod\_{i=r+1}^{m}(x-\lambda\_i)$$ admits exactly $(m+r+\mathrm{deg}(S))$ real roots pairwise distinct. 


In other words I am looking for a real polynomial $P$ which satisfies
$$P(λ\_i)=μ\_i,1≤i≤m \mbox{ and } P'(λ\_i)=0,1≤i≤r,$$
 and whose roots are all real and simple.



A special polynomial interpolation
If there are no restrictions on $S$, one can apply the following construction.
Let $$Q(x)=S(x)\prod\_{j=1}^r(x-\lambda\_j)^2\prod\_{j=r+1}^m(x-\lambda\_j).$$
We want to construct $S$. Consider first the case when $R$ does not change sign on the real
line. Suppose it is positive. Then choose $S$ so that 


a) All double zeros of $Q$ are maxima. This is achieved by inserting zeros of $S$ at properly chosen places.


b) That the minima of $Q$ on each interval between its zeros are less
(more negative) than $-\max R$
(the maximum of $R$ is taken over some big interval that contains all zeros of $Q$).
This can be achieved just by multiplying $S$ on a sufficiently large constant.


Now when you graph $-R$ and $Q$ on the same picture, you see that the graphs intersect as many
times as the degree of $Q$ is. So $R+Q$ has all zeros real. You also make them simple,
by another multiplication of $Q$ on a constant.


The general case is similar. Consider the real zeros of $R$, say $x\_1,...,x\_k$.
On the intervals $(x\_k,x\_{k+1})$ where $R(x)>0$, you want all double zeros of $Q$ to be maxima.
On the intervals $(x\_k,x\_{k+1})$ where $R(x)<0$, you want all double zeros of $Q$ to be minima.
This is achieved by placing the zeros of $S$ at appropriate places. Each new zero of $S$
switches the sign of $Q$ at this place.


Now after you achieved this (that all double roots of $Q$ are maxima where $R$ is positive, and
minima where it is negative), it only remains to multiply $Q$ by a very large positive constant,
to make sure that the graphs of $-R$ and $Q$ intersect as many times as needed.
Finally, by slightly perturbing the constant multiple in $S$, you achieve that all
these real roots are simple.


But I don't now how useful this polynomial can be, as its degree can be larger than the number
of interpolation nodes. You can try to minimize the degree of $S$, the answer will depend
on the relative position of the double roots $\lambda\_j$ among all $\lambda\_j$.
But roughly speaking the degree of $Q$ constructed with this method, can be about twice as large as that of $R$. 


