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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 $a_n x^n + g(x)$?

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    $\begingroup$ The polynomials $f(x)=1$, $g(x)=2$ and $h(x)=x^2+1$, $k(x)=x^2+x+1$ have the same roots over $\Bbb R$. $\endgroup$ – ajotatxe Jun 6 at 16:15
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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.

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  • $\begingroup$ Aha okay, thanks. But then how come you can write a polynomial in terms of its roots? Like $(\lambda - a)(\lambda - d)-bc = 0$ can be written in terms of its roots $(\lambda - \lambda_1)(\lambda - \lambda_2) = 0$? Since having the same roots apparently does not imply that two polynomials are identical, using the roots as a way to write a unique polynomial then seems confusing to me $\endgroup$ – Fac Pam Jun 6 at 16:36
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    $\begingroup$ @FacPam Those polynomials are monic, and since they are quadratic, there are always exactly two (possibly complex) roots when counted with multiplicity. And since they both have the two roots $\lambda_1$ and $\lambda_2$, they do turn out to be the same polynomial. Wasn't this addressed in your previous question? $\endgroup$ – Arthur Jun 6 at 16:40
  • $\begingroup$ Ahh, no it was not addressed - at least I do not think so. Possibly because I do not know the definition of "monic". I will look that up now. Thanks again $\endgroup$ – Fac Pam Jun 6 at 16:48
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    $\begingroup$ @FacPam "Monic" just means that the highest-order term has coefficient $1$. That requirement is there to stop things like the first counterexample in my answer: $x^2-1$ is monic (as the coefficient of $x^2$ is $1$) while $2x^2-2$ is not (as the coefficient of $x^2$ is not $1$). $\endgroup$ – Arthur Jun 6 at 17:14
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    $\begingroup$ @FacPam No, the "same multiplicities" is to resolve the issue of $x-1$ versus $x^2 - 2x + 1$ (or, as others have pointed out, $x$ versus $x^2$). They have the same roots, but are clearly not the same. That's because these quadratic example polynomials have a double root, a root with multiplicity $2$. $\endgroup$ – Arthur Jun 6 at 18:47
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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.

enter image description here

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.

enter image description here

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    $\begingroup$ How about spicing it up with a polynomial for which (X-2) appears twice? Still the same roots, and it would show that we're talking about more than just trivial constant factors. $\endgroup$ – Andras Deak Jun 7 at 15:38
  • $\begingroup$ Remind your students that linear functions are polynomial, and show them $f(x)=x,\, g(x)=-x, $ and $h(x)=2x.$ $\endgroup$ – DanielWainfleet Jun 10 at 6:28
  • $\begingroup$ I think you meant to put this comment on a different answer. $\endgroup$ – JoeTaxpayer Jun 10 at 10:23
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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 $f \ne g$.

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    $\begingroup$ Of course for finite fields $\mathbb{F}$, the pigeonhole principle alone can say that there will be distinct polynomials which induce the same map $\mathbb{F}\to\mathbb{F}$. Because the number of such maps is finite, while the number of polynomials is infinite. $\endgroup$ – Jeppe Stig Nielsen Jun 8 at 8:27
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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 $f_4(x)=x$.

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  • $\begingroup$ What do you mean by "up to a constant" - $f_1(x)=(x+1)(x-2)$ and $f_2(x)=5(x+1)(x-2)$ are not identical? $\endgroup$ – Fac Pam Jun 6 at 20:29
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    $\begingroup$ @FacPam Well, $f_2(x) = 5 f_1(x)$ so we can hardly say they are identical except at their roots $\endgroup$ – zdimension Jun 7 at 6:51
  • $\begingroup$ @FacPam It means that if $f(x_o)=0$ for some $x_0$, then also $\lambda f(x_0)=0$ for any scalar $\lambda$ $\endgroup$ – Tesla Jun 7 at 8:31
  • $\begingroup$ if you don't consider multiplicity you first stmt is incorrect, consider x(x-1)^2 and x^2(x-1). If you consider multiplicity, then your second stmt incorrect $\endgroup$ – RiaD Jun 7 at 9:15
  • $\begingroup$ yea thanks all, didnt think about it for more than two seconds. $\endgroup$ – Tesla Jun 8 at 6:08
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The multiplicity counts too: for example $x$ and $x^2$ have the same roots, but are different polynomials. If two polynomials have all the same roots and all the same multiplicities, then yes they are equal. However, this statement needs to be interpreted correctly: you need to work over $\mathbb{C}$ (or some other algebraically closed field). For example, over $\mathbb{R}$, the polynomials $x^2+1$ and$ (x^2+1)^2$ have the same real roots (namely, they have no roots!) but are clearly not the same.

So: you have to count the roots with multiplicity in the algebraic closure.

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No. You probably wouldn't consider $f(x)=x$ and $f(x)=10x$ to be identical even though they have the same root.

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Let's start by considering polynomials with all their roots, real and complex. This allows us to fully answer the question for first complex, and then real, polynomials and roots.

This approach will not only let us get all answers, but prove these are all answers, and the only answers.** It's also easy to see why that's so.

Fundamental principle: Over the complex numbers, all nonconstant polynomials can be uniquely factored into linear terms and a multiplier

See Wikipedia "Irreducible polynomial - over the complex numbers" and Fundamental theorem of algebra: any nonconstant polynomial can, in complex terms, be uniquely factored into something like

A.(x-B).(x-C).(x-D)... = 0

A <> 0 and B, C,D.. are the roots. B,C,D can of course be complex or real numbers. Also some of the B, C, D... may repeat, in which case we have one or more repeating roots, but the polynomial will still factorise this way.

We can rewrite this in terms of unique roots, as follows:

A. [(x-B)^P] . [(x-C)^Q] . [(x-D)^R] . [...] ... = 0

where A <> 0 and B,C,D... are now all unique complex numbers, and are the roots of the polynomial, and P,Q,R... are all integers >= 1 that account for any repeated roots.

The fundamental.theorem of algebra guarantees we can factor all polynomials this way, and that it will be unique for each polynomial. It's also evident from inspection that B,C,D are the roots, and all the roots, and no other roots exist.

Your answer, if complex roots are allowed

... Is now quite simple. Suppose 2 non-constant polynomials have identical roots. Then they must be identical other that possibly:

  • a different non-zero multiplier (A is different between the polynomials, when factored)
  • repeated roots (one or more of P, Q,R will differ between the polynomials, when factored)

What if we only allow real roots?

The polynomial can still only be factored one way as above. The only difference is, any B,C,D that isn't a real number won't ever equal a value of X we can choose, so it can't be a solution. So as well as the 2 types of change above, we can also change the powers for any existing complex linear factors to any integer >= 0, or multiply by new complex linear factors (to any integer power >0), and provided the factor we multiply/divide by has a complex parameter, it won't ever affect the real roots. We can't divide by new complex linear factors, though, because the result wouldn't be a polynomial.

This is easiest explained by example.

Example: suppose our equation is a polynomial that factors into a mix of real and complex linear factors, some repeated:

4 . (X - 7)^2 . (X + 4.5) . (X + 2i) . (X - 2i)= 0

Then any polynomial with identical real roots must be formed by some combination of these changes (I'll give an example of each):

  • (-6) . (X - 7)^2 . (X + 4.5) . (X + 2i) . (X - 2i) = 0
    We have multiplied A by some real value <> 0 (in this case, -1.5).

  • 4 . (X - 7)^8 . (X + 4.5) . (X + 2i) . (X - 2i)= 0
    4 . (X - 7)^0 . (X + 4.5)^5 . (X + 2i) . (X - 2i)= 0
    We have changed the powers for some of the repeated roots (up or down)

  • 4 . (X - 7)^2 . (X + 4.5) . (X + 2i) . (X - 2i) . (X - [3+7i])^3 = 0
    4 . (X - 7)^2 . (X + 4.5) . (X + 2i)^17 . (X - 2i) = 0
    4 . (X - 7)^2 . (X + 4.5) . (X + 2i)= 0
    4 . (X - 7)^2 . (X + 4.5) = 0
    We have changed the powers for some of the complex roots (up or down), or removed them (equivalent to changing their power to 0), or introduced new complex linear factors.

Note that this last transformation might or might not change some of the coefficients in the equation from real to complex coefficients or vice-versa, depending what you do (see especially the last example where they don't). It may well change the complex roots of the polynomial. But it will not change, add or remove any real roots of the polynomial.

If you restrict yourself to changes of this kind that don't change any real coefficients to complex coefficients, you'll achieve all real coefficient polynomials with the same roots this way.

** Note - For quintics and higher, we may not be able to factorise to simple algebraically expressed roots, because not all 5th and higher order polynomials allow for neat expressions of their roots this way. But - even if inexpressible - the roots do exist, the limitation is in our ability to calculate them exactly, or write them concisely, not in their existence. The same method will work and be valid, and the same other types of polynomials will have identical complex (or real) roots. We just wouldn't be able to calculate or write the linear expressions, transformative equations, or roots, neatly, in the same way.

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    $\begingroup$ please learn to use MathJax $\endgroup$ – qwr Jun 7 at 14:03
  • $\begingroup$ This answer should really be upvoted $\endgroup$ – klutt Jun 9 at 18:57
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Of course NOT. A simple multiplication by a constant works. More interestingly define an equivalence relation where p1~p2 iff they share exactly the same roots!

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