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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 $g(x)$?

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  • $\begingroup$ 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 $f(x)=a_2(x - r_1)(x - r_2)$. $\endgroup$ – NoChance Jun 7 '19 at 11:38
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By the fundamental theorem of algebra every polynomial can be written as the product of linear factors. These give you exactly the zeros and their multiplicity. So yes, every monic polynomial has a unique set of roots with respect to their multiplicity.

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.

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  • $\begingroup$ 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 $2$? $\endgroup$ – Fac Pam Jun 7 '19 at 10:42
  • $\begingroup$ Every polynomial has a unique set of roots with respect to multiplicity. Every set of roots has a unique monic polynomial. $\endgroup$ – Floris Claassens Jun 7 '19 at 10:57
  • $\begingroup$ 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 :) $\endgroup$ – Fac Pam Jun 7 '19 at 11:06
  • $\begingroup$ 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. $\endgroup$ – Floris Claassens Jun 7 '19 at 11:51
  • $\begingroup$ @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 $a_n\neq0$. $\endgroup$ – LegionMammal978 Jun 9 '19 at 0:39

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