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 $g(x)$.

Question

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 $f(x)$?

Answer 1

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 $g$.

Answer 2

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 $\ddot\smile$