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Is there a way to find all roots of a polynomial equation?

Lets say $$x^5+ax^4+bx^3+cx^2+dx+e=0$$

how to find its all roots?

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    There are a number of approximation methods for finding solutions of polynomial equations. However, there is no formula akin to the quadratic formula for finding the roots of arbitrary polynomials of degree $5$ of higher in terms of the coefficients, using only addition, multiplication, division, and root extraction.2012-04-18
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    You want this in symbolic form, in terms of $a,b,c,d,e$ ?? Using only the usual algebraic operations ?? Then the answer is: NO, there is no way.2012-04-18
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    Depends what you mean by find. Given the numbers $a$ to $e$, we can *approximate* the roots arbitrarily closely.2012-04-18
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    Luna: Celebrated mathematician [Niels Abel](http://es.wikipedia.org/wiki/Niels_Henrik_Abel) proved there is no closed form for an arbitrary polynomial of degree $5$ or grater. Thus, as other users are suggesting, any solution will be a "mere" approximation.2012-04-18
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    It you mean solvability by radicals, only the linear degree, quadratic, cubis and quadratic equations are solvable. The general quintic or higher degree equations are not solvable by radicals.2012-04-18
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    Going beyond radicals: http://en.wikipedia.org/wiki/Quintic_function#Beyond_radicals2012-04-18
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    You must put specifications on a,b,c,d, e to get accurate responses. For example, if e=0, it is reduced to a 4-th degree equation.2012-04-18
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    @EmmadKareem: The coefficient of $x^5$ is $1$, even if $e = 0$...2012-04-18
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    @Aryabhata, you are correct but when $e=0$, we have $x^5+ax^4+bx^3+cx^2+dx=0$ which can be reduced to 4ht degree by dividing by x (for non-zero x).2012-04-18
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    @EmmadKareem: Yes, but you are still solving a quintic :-) Saying that OP "_must_ put specifications to get accurate response" is imposing conditions for no reason, without even knowing what OP is really asking. For instance, "find roots" could just mean some numerical method. And if you see some links in the comments (to the question and one of the answers), you don't really need restrictions for a "closed form" accurate solution. Besides, OP might not have any specifications to add.2012-04-18
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    @Aryabhata, you have a point. Thanks for the explanation.2012-04-18
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    possible duplicate of [Is there a general formula for solving 4th degree equations?](http://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations)2012-04-19

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A theorem by Abel and Ruffini (see e.g. http://en.wikipedia.org/wiki/Abel-Ruffini_theorem) states that there is no general way of expressing (explicitly) the roots of a polynomial of order 5 or more or said differently, that there exists polynomials of order 5 or more for which it is impossible to do so. (However, and as was mentioned in other comments, there exists ways of approximating the roots with arbitrary precisions)

Now there exists methods to find the number of roots in a particular zone. See e.g. Sturm method, Budan-Fourier, Routh-Hurwitz (argument principle),

and also, exclusion/inclusion theorem (e.g. van der Sluis' theorem, Laguerre's theorem).

So the brief answer to your question is: no there is not. However there are a lot of methods to characterize the localization of the roots or to approximate the roots.

EDIT: precisions added following comments by Franklin.vp.

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    The theorem by Abel and Ruffini doesn't say that.2012-04-18
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    There are "closed form" solutions which don't involve radicals (for instance, see here:http://en.wikipedia.org/wiki/Quintic_function#Beyond_radicals). So, this answer is misleading (it never mentions radicals!). And, "finding the roots" is a pretty overloaded term. What exactly does it mean?2012-04-18
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    http://en.wikipedia.org/wiki/Quintic_function#Beyond_radicals2012-04-18
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    Put another way: yes, the roots of polynomials of degree higher than four can be expressed in closed form, but they involve the use of special functions, like hypergeometric, elliptic, or theta functions. This is completely analogous to the *casus irreducibilis* of the cubic equation, which requires trigonometric/hyperbolic functions for the solution...2012-04-19
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I'm reading this paper of AMS http://www.ams.org/bookstore/pspdf/stml-35-prev.pdf really, an equation of degree 5, is not easy to solve, unless that you have some properties over the coefficients.