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Any real polynomial can be expressed as a product of quadratic and binomial factors like $(x+a)$ and $(x^2 + bx + c)$. Given a polynomial, is there an algorithm which will find such factors?

For example, how can I express $x^4 +1$ in the form $(x^2 + bx + c)(x^2 +dx + f)?$

Mark
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    You're looking for numerical approximations, right? A computer can't store all the digits of a transcendental, for instance. – anon Mar 01 '13 at 01:20
  • Maybe heuristic is a better word than algorithm. The motivation is that many test questions will ask you to integrate rational functions and the strategy is to factor the denominator in order to decompose into partial fractions. – Mark Mar 01 '13 at 01:30
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    In general there is no heuristic. Arbitrary real-coefficient polynomials can even have linear and quadratic factors involving numbers that cannot be expressed in terms of addition, multiplication, radicals or other elementary operations. However, your comment makes me suspect you are talking specifically about rational-coefficient polynomials, of which we can talk about paper-and-pen methods to factor fully into quadratic and linear factors. – anon Mar 01 '13 at 02:37
  • Yes I guess I should have clarified. I am interested in polynomials with rational coefficients. Is there a general approach? – Mark Mar 01 '13 at 03:30
  • I found this helpful question: http://math.stackexchange.com/questions/68178/finding-roots-of-polynomials-with-rational-coefficients – Mark Mar 01 '13 at 03:44

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If there were a simple algorithm for this, for whatever definition of simple, then there would be a simple algorithm for solving polynomial equations of arbitrary degree, and no such algorithm exists (for a precise definition of simple here).

For the example you gave, you can try the method of undetermined coefficients: just expand the product and equate coefficients.

lhf
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Well, there is the Lin-Bairstow Method. It uses numerical approximations based on Newtons Method to determine the coefficients of quadratic factors. Very effective.

Factoring arbitrary polynomials over the real? Notice you said reals, not rationals. I cannot conceive of an algorithm not relying on numerical approximations that can achieve this, since the coefficients are going to be arbitrary real values (potentially irrational). And if you allow for that, then the Lin-Bairstow method is sufficient to solve your problem.

The reason why this method is preferable is because it is a general method to find any and all quadratic factors over real coefficients, thereby allowing you to solve even for complex solutions. Complex solutions will be the roots of these general quadratic factors. All achieved using a reiterative numerical method on real numbers; perfectly achievable on real number calculators. I find that fascinating, personally.

For the special case of cubics, quartics, and a select few quintics, there are algebraic approaches, but no general algebraic solution exists for 5th degree and higher polynomials.

CogitoErgoCogitoSum
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$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt2x)^2$ and so on.

Gerry Myerson
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If you can find roots, you can find factors. As others have pointed out, this needs to be done using numerical methods for polynomials with degree greater than 4. In fact it's often a good idea to use numerical methods (rather than closed-form formulae) even in the degree 3 and degree 4 cases, too. There's lots of good software for numerically finding roots of polynomials. The best-known is the Jenkins Traub method

bubba
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  • Actually, the second variant of Jenkins-Traub, the one for real polynomials, finds real quadratic factors just like Bairstows method. Despite being the more complicated variant, it is the most copied one. – Lutz Lehmann Dec 17 '13 at 10:03