Are polynomials with the same roots identical?

I know that polynomials can be refactored in terms of their roots. However, this must imply that two different polynomials have different roots (this is just what I think). So my question is: Are polynomials with the same roots identical? - if so, why?

A follow-up question that is also about the uniqueness of roots and polynomials can be found here: Is the set of roots unique for each $g(x)$ in $a_n x^n + g(x)$?

Solution 1:

No, they are not.

For instance, $2x^2-2$ and $x^2-1$ have the same roots, yet they are not identical.

And, depending on what you mean by "the same roots", we have that $x^2-2x+1$ and $x-1$ have the same roots, yet they are not identical.

Again, depending on what you mean by "the same roots", $x^3+x$ and $x^3+2x$ both only have one real root, yet they are not the same.

However, if two monic polynomials have the same roots, with the same multiplicities, over some algebraicaly closed field (like the complex numbers $\Bbb C$) then yes, they are identical.

Solution 2:

The accepted answer is deservedly so, a great explanation. As I read this, I thought of my students who are visual learners, for whom, a picture is worth a thousand words, and this would answer their question with almost no further comment.

The image above shows a simple $Y=(X-1)(X-2)(X-3)$ and an overlapping $Y=-3(X-1)(X-2)(X-3)$.

This helps show that manipulation made to an equation such as factoring may preserve the roots, but do not leave an equation with the the same nature, e.g. the end behaviour which might be important, is easily lost.

Edit - by popular demand, I'm adding the original graph, and an overlapping one with 2 as a double root.

Solution 3:

For polynomials over $\mathbb{R}$, the answer is no; for example, $f(x)=x$ and $g(x) = x(x^2+1)$ have the same roots over $\mathbb{R}$—with the same multiplicities—but they are not equal.

For polynomials over $\mathbb{C}$, the answer is almost. The fundamental theorem of algebra says that every polynomial over $\mathbb{C}$ of degree $n \ge 1$ splits uniquely into $n$ linear factors. So if $f$ and $g$ have the same roots $\alpha_1,\alpha_2,\dots,\alpha_n$, listed with multiplicity, then $$ f(x) = \lambda (x-\alpha_1)\cdots(x-\alpha_n) \text{ and } g(x) = \mu(x-\alpha_1)\cdots(x-\alpha_n) $$ for some $0 \ne \lambda,\mu \in \mathbb{C}$. So roots (with multiplicity) determine polynomials over $\mathbb{C}$ up to a multiplicative constant and, in particular, monic polynomials over $\mathbb{C}$ are uniquely determined by their roots.

For polynomials over finite fields, the answer is very much no. There are polynomials that don't just have the same roots, but they have all the same values for every input. For example, the polynomials $f(x) = x$ and $g(x)=x^3$ over $\mathbb{F}_2$ satisfy $f(x)=g(x)$ for all $x \in \mathbb{F}_2$, and yet $f \ne g$.

Solution 4:

No, they aren't:

$f_1(x)=(x+1)(x-2)$ and $f_2(x)=5(x+1)(x-2)$ have the same roots. But they don't even need to have same degree to have the same roots: $f_3(x)=x^2$ has the same root as $f_4(x)=x$.