# Is the set of roots unique for each $g(x)$ in $a_n x^n + g(x)$? [duplicate]

I woke up and suddenly wondered about the following: A polynomial of degree $$n$$ $$a_n x^n + a_{n-1} x^{n-1} + \cdot\cdot\cdot + a_{n-n} x^{n-n}$$ can be written as $$a_n x^n + g(x)$$ where $$g(x)$$ is a polynomial of degree $$n-1$$. Then I was thinking that for every $$g(x)$$ has a unique set of roots with respect to their values and multiplicities - is this true?

So for instance consider a polynomial of degree $$2$$: $$x^2 + a_1 x + a_0$$ which can be written as $$x^2 + g(x)$$ where $$h(x) = a_1 x + a_0$$. Then every unique combination of $$a_1$$ and $$a_0$$ in $$g(x)$$ will give a unique set of roots in terms of values and multiplicities.

An example of this is $$x^2 + 5x + 4$$ (with $$a_2 = 1$$) - this polynomial has the roots $$-4$$ and $$-1$$, both with a multiplicity of $$1$$. Thus, for this $$a_2$$, only $$g(x) = 5x + 4$$ will give this set of roots. Only for a different $$a_n$$ a different $$g(x)$$ may yield this set of roots. I mean after all, any polynomial of degree $$2$$ can be written as $$a_2(x - r_1)(x - r_2)$$.

This is all something that I think could be true. So can any of you smart guys out there tell me if I am right and if so, might explain why this holds - that the set of roots with respect to their values and multiplicities are unique for each $$g(x)$$?

• You are saying "I mean after all, any polynomial of degree $2$ can be written as $a_2(x - r_1)(x - r_2)$." If $r1,r2$ are roots you can write the function as $f(x)=(x - r_1)(x - r_2)$ NOT as $f(x)=a_2(x - r_1)(x - r_2)$. – NoChance Jun 7 '19 at 11:38

i.e. if we let $$g(x)$$ be a polynomial of degree $$n-1$$, by the fundamental theorem of algebra we can find a unique $$c\in\mathbb{C}$$ and unique $$x_{1},...,x_{n-1}\in\mathbb{C}$$ such that $$g(x)=c(x-x_{1})(x-x_{2})...(x-x_{n-1}).$$ Hence the set of roots of a polynomial is unique for said polynomial up to multiplication by a scalar.
• But monic just means that the leading coefficient is 1 - so isn't this just one case for which the set of roots for $g(x)$ is unique for this leading coefficient $a_n$. Wouldn't the uniqueness also hold when the leading variable $a_n \neq 1$, like $2$? – Fac Pam Jun 7 '19 at 10:42
• So it is true that the set of roots with respect to their values and multiplicities are unique for each $g(x)$ in $a_n x^n + g(x)$? Not only for $1 x^n + g(x)$ (monic) but any $a_n x^n + g(x)$. It would be very nice if you could elaborate :) – Fac Pam Jun 7 '19 at 11:06
• I've expanded my answer. I am not sure what the value is of considering $a_{n}x^{n}+g(x)$. In you're question you are only concerned by the roots of $g(x)$ and for this it only matters that $g(x)$ is a polynomial. – Floris Claassens Jun 7 '19 at 11:51
• @FacPam Take $g^\prime(x)=\frac{g(x)}{a_n}$ and you're back to your original $1x^n+g^\prime(x)$. This would work for any particular $a_n\neq0$. – LegionMammal978 Jun 9 '19 at 0:39