Let $f(x)$ and $g(x)$ be two monic polynomials of the same degree such that adding $1$ to the roots of $f(x)$ we get the roots of $g(x)$. Let $f(x)$ and $g(x)$ be two monic polynomials of the same degree such that adding
$1$ to the roots of $f(x)$ we get the roots of $g(x)$. Then does their any relations between the constant term of $g(x)$ and the constant term of f(x).
i.e I want to say that can we find the constant term of $g(x)$ from the constant term of $f(x)$? 
 A: It depends on whether adding 1 gives you the roots with multiplicities, or whether the multiplicities don't have to match up. So here's an example:
Suppose $f(x) = (x-1)^{2} (x- 2)^{3}.$ Then, we have a few choices for $g(x)$:
$g(x) = (x-2)^{2}(x - 3)^{3}$, $g(x) = (x-2)^{3} (x - 3)^{2}$.
In both cases, the set of roots is $\{1 + \alpha : \alpha \text{ is a root for $f$}\}.$ But they have different constant terms.
However, if you preserve multiplicities, then you can get the entire polynomial $g$ from $f$:
Over the complex numbers, we can factor $f$ into a product of linear polynomials: 
$$f(x) = \prod_{i=1}^{n} (x - \alpha_{i})$$
Then, if we preserve multiplicities of the roots and just add 1 to them, this tells us that 
$$g(x) = \prod_{i=1}^{n} (x - \alpha_{i} - 1).$$
So to answer your question in particular, if $a_{0}$ is the constant term in $f$ and $b_{0}$ is the constant term in $g$, then
$$a_{0} = (-1)^{n} \prod_{i= 1}^{n}  \alpha_{i}, b_{0} = (-1)^{n} \prod_{i=1}^{n} (\alpha_{i}  +1).$$
So, you can't get the constant term of $g$ just by knowing the constant term of $f$, but there are some relations. If you know all of $f$, then you can get all of $g$.
A: Consider:


*

*The following two monic polynomials have the same degree and the same constant term:
\begin{align}
(x-3)(x-4) &= x^2-7x+12,\\
(x-2)(x-6) &= x^2-8x+12,
\end{align}

*but the relevant shifts have different constant terms:
\begin{align}
(x-4)(x-5) &= x^2-9x+20,\\
(x-3)(x-7) &= x^2-10x+21.
\end{align}


I hope this helps $\ddot\smile$
