Let $\lambda\_1,\ldots,\lambda\_m$ real numbers pairwise distinct and $\mu\_1,\ldots,\mu\_m$ real numbers all nonzero.
We know from the Lagrange polynomial interpolation that there exists an unique polynomial $R$ of degree less than $(m+1)$ such that $R(\lambda\_i)=\mu\_i,1\leq i\leq m$. We can prove by the means of an adequate use of the intermediate value Theorem, the existence of a real $\alpha$ such that the polynomial $P\_{\alpha}(x)=R(x)+(x-\alpha)(x-\lambda\_1)\ldots(x-\lambda\_m)$ admits $(m+1)$ real roots pairwise distinct.


I wonder if it is possible to generalize this property. Precisely, let $1\leq r\leq m $, we denote by $Q$ the polynomial of degree less than $(m+r+1)$ such that $Q(\lambda\_i)=\mu\_i,1\leq i\leq m$ and $Q'(\lambda\_i)=0,1\leq i\leq r$. Is it possible to find $(\alpha,\beta)\in\mathbb{R}^2$ such that the polynomial
$$S\_{\alpha,\beta}(x)=Q(x)+(x-\alpha)(x-\beta)\displaystyle\prod\_{i=1}^{r}(x-\lambda\_i)^2\displaystyle\prod\_{i=r+1}^{m}(x-\lambda\_i)$$ admits $(m+r+2)$ real roots pairwise distinct?



Polynomial interpolation whose roots are real and simple
The answer is no. (Assuming that $Q$ is indeed the unique interpolation polynomial of degree $m-1$).
Indeed, take all your $\mu\_j=1$. then $Q$ will be identically equal to $1$.
Then choose your other data $\lambda\_j$ in such a way that the polynomial
$$p(x)=\prod\_{j=1}^r(x-\lambda\_j)^2\prod\_{j=r+1}^m(z-\lambda\_j)$$
has the following shape: when you look at the graph from left to right, it has a series (say, more than 3) of multiple zeros,
then a single simple zero, then a series of multiple zeros again, and a simple zero, and again
a series of multiple zeros.
All zeros are real. Than you see that on each series of multiple zeros, $p^{\prime\prime}(\lambda\_j)$
has the same sign, and this sign alternates: one sign in the first series, opposite sign in
the second series, and so on.


Multiplying by a second degree factor $(x-\alpha)(x-\beta)$, 
will alter this signs pattern somewhat, but not too much.
In any case some double root with POSITIVE second derivative will remain. This root will
be a local minimum of $(x-\alpha)(x-\beta)p(x)$.


Now it is evident that when you have an equation of the form $1+q(x)=0$, where
$q$ is a polynomial with all roots real, and one of them $x\_0$ is of multiplicity $2$, and the $p^{\prime\prime}(x\_0)>0$,
then this equation must have some non-real roots.
Just graph on the real line to see it.


