Let $a,d$ be polynomials of $\mathbb Z[X]$ with $\deg a>\deg d\ge0$ and $P$ be a polynomial of $\mathbb Z[X]$. Consider an infinite sequence of integers $(\lambda\_n)\_n$. Can one assert there exists a $\lambda\_n$ such that 
$$(\lambda\_n +1)^2P-(\lambda\_n a+d)(a+\lambda\_n d)$$ has only simple roots?


EDIT: The polynomials $a$ and $d$ are relatively prime in $\mathbb Q[X]$.


Thanks in advance.



Polynomials with no multiple root
It can happen that $(\lambda+1)^2P-(\lambda a+d)(a+\lambda d)$ has multiple roots for any integer $\lambda$: namely, if $P=ad$, then
\begin{multline\*} 
 (\lambda+1)^2P-(\lambda a+d)(a+\lambda d) \\ 
 = (\lambda^2+1)(P-ad) + \lambda(2P-a^2-d^2)
 = -\lambda (a-d)^2, 
\end{multline\*}
so that any root of $a-d$ is a multiple root of $(\lambda+1)^2P-(\lambda a+d)(a+\lambda d)$.




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### A little beyond


One can actually classify completely the cases where $\lambda\_n$ with the property in question can be found. Namely, from the identity
 $$ (\lambda+1)^2P-(\lambda a+d)(a+\lambda d)
 = (\lambda+1)^2(P-ad)-\lambda(a-d)^2 $$
it follows that if some root of $a-d$ is a multiple root of $P-ad$, then the polynomial in the left-hand side has multiple roots *for any integer $\lambda$*. On the other hand, if none of the roots of $a-d$ is a multiple root of $P-ad$, then there are only finitely many those $\lambda$ for which the polynomial has multiple roots; this follows from the general fact that if the polynomials $P$ an $Q$ do not have common multiple roots, then there are at most finitely many $\lambda$ for which $P+\lambda Q$ has a multiple root.


