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$$f(x)=a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0$$

and let the roots of the polynomial be arranged from lowest to highest: $r_1 \leq r_2 \leq \cdots \leq r_n$.

Then the list of the all roots of polynomial $f(x)$ can be express in to new polynomial, let's say $$g(x)= (1\text{st},r_1),(2\text{nd},r_2)\cdots(n\text{th},r_n).$$ Example: Starting with $f(x)=x^2-8x+15$ or $(x-3)(x-5)$, then $r_1=3,r_2=5$ now when $x=1,g(x)=3$ and $x=2,g(x)=5$. Therefore $g(x)=2x+1$.

Practical application: If you find one of the roots of polynomial therefore the remaining roots of $f(x)$ can be solved. Find the general equation of $g(x)$ where $f(x)=0$ http://www.wolfram.com/technology/guide/GigaNumerics/

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I don't follow you. –  Cocopuffs Jun 17 '12 at 20:59
I don't see a question here. –  Qiaochu Yuan Jun 20 '12 at 7:21

2 Answers 2

It looks like you want a polynomial $g$ such that $$g(1)=r_1,\; g(2)=r_2,\;\ldots,\; g(n)=r_n$$ where $r_1,\ldots,r_n$ are the roots of the polynomial $f$. There are infinitely many polynomials that have this property, for any choice of numbers $r_1,\ldots,r_n$ whatsoever; but you can use Lagrange interpolation to find the polynomial of least degree that does this.

However, since there need not be a nice expression for the roots $r_1,\ldots,r_n$ of $f$ in terms of the coefficients $a_n,\ldots,a_0$, there will not be a nice expression for the interpolating polynomial $g$ in terms of the coefficients, either.

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If it were possible to do what you want, then you could take any polynomial whatsoever, multiply it by $x$ to get a polynomial for which you know one of the roots (namely, zero), then apply your technique to the new polynomial to get all the roots of the original polynomial. That is, you'd have a way of solving any polynomial whatsoever.

Considering how long people have been trying to sove polynomials, it seems unlikely that everyone would have missed a trick like this, don't you think?

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