Smoothness of the sum and product of paired roots (eigenvalues) I'm a little upset to ask the following question which might be naive, but I'm really not familiar with such topics and need help.
Consider a parameter-dependent polynomial equation
\begin{equation}
 a_n(p)\lambda^n + a_{n-1}(p)\lambda^{n-1} + \cdots + a_2(p)\lambda^2
+a_1(p)\lambda + a_0(p) = 0,
\end{equation}
where $p\in\mathbb{R}^m$ is the parameter, and all $a_i(p)$ are
sufficiently smooth in $p$. Suppose that at some critical parameter value, say,
$p^*=0$, we have
\begin{equation}
a_1(0)=a_0(0)=0, ~~a_2(0)\neq0, a_n(0)\neq0
\end{equation}
so the polynomial has a double-zero root.
Let $\lambda_1(p)$ and $\lambda_2(p)$ be the two roots,
continous in $p$, such that $\lambda_1(0)=\lambda_2(0)=0$.
My question is, is it guaranteed that the functions
$\lambda_1(p)+\lambda_2(p)$ and $\lambda_1(p)\lambda_2(p)$ are smooth
in $p$, at least in a neighborhood of $p=0$?
Intuitively I thought this might be true, but I don't know how to confirm
it or give any counterexamples to deny it. Can someone enlight me on this?
Thanks in advance.

The remainder under polynomial division of a polynomial $a(λ)$ by a quadratic term $b(λ)=λ^2+b_1λ+b_0$ is a smooth function as it is  polynomial in the coefficients. The derivative of this 2D valued function $r(λ)=a(λ)\mod b(λ)$ for the coefficients $b_0,b_1$ is regular if the two factors $b(λ)$ and $c(λ)$ in $a(λ)=b(λ)c(λ)+r(λ)$ are relative prime.
In the given situation the factorization at $p=0$ with the quadratic factor $λ^2$ is relative prime as $a_2(0)\ne 0$. Thus one can by the implicit function theorem the quadratic factor locally to $b(p,λ)=λ^2+b_1(p)λ+b_0(p)$ so that it really is a factor of $a(p,λ)$. The continuation is analytical as solution of an algebraic system.
By Viete, the product of the roots is $b_0(p)$ and the sum of the roots $-b_1(p)$.
[See Bairstow method for a loosely related numerical method.]
