For any $m,n\in \mathbb{N}$ there exists a polynomial $p(x)$ such that $p(x)+1$ has $m$ negative roots and $p(x)-1$ has $n$ negative roots .

Question

I want to find a polynomial $p(x)$ passing through origin such that $p(x)+1$ has $m$ negative zeroes and $p(x)-1$ has $n$ negative zeroes for arbitrary $m,n$ ??

I observed graphically that it is possible but don't know exact way to solve this .