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Factoring methods with equations higher than second degree

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Can someone tell what are the other factoring methods when dealing with equations higher than the second degree. The only I is the synthetic division. Aside from that, I do not know anything. I need other factoring methods so I can deal with problems that cannot be factored using synthetic division. I will use them in solving the unknowns in determinants. Please explain briefly to me and give such example.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 7, 2010 #2

    HallsofIvy

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    Re: factoring

    There are no general methods of factoring such polynomials. The most helpful thing often is the "rational root theorem". If a polynomial is of the form [itex]a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0[/itex], with all coefficients integers, then any rational number that makes it 0 will be of the form p/q where q is an integer that evenly divides [itex]a_n[/itex], the leading coefficient, and p is an integer that evenly divides [itex]a_0[/itex], the constant term. If [itex]a_n[/itex] and [itex]a_0[/itex] don't have too many factors, you could try all combinations to see if any give a rational root. If there exist a rational root, p/q, then you know that (x- p/q) is a factor. Of course, it might happen that there are NO rational roots and so all factors will have be irrational or complex. In that case, there are not simple ways to find them for degree higher than 2. There exist complicated formulas for degrees 3 and 4 but it was shown in the nineteenth century that polynomial equations of degree 5 and higher may have solutions that cannot be written in terms of roots so there cannot be any formulas, in terms of roots and other algebraic operations, for them.
     
  4. Feb 7, 2010 #3
    Re: factoring

    so what should I do if the the given cannot be factored by synthetic division? I already encountered one in differential equations and I do not know what to do.
     
  5. Feb 7, 2010 #4
    Re: factoring

    As HallsofIvy pointed out, there are no general methods, except if you have some additional information (like e.g. you know that the roots are rational). Besides the Rational Roots Theorem, there are then other sophisticated methods like doing computations modulo some prime and then using Hensel lifting to convert that result to modulo some large power of that prime. You can use so-called rational reconstruction methods to convert integers modulo large numbers back to fractions.

    https://openaccess.leidenuniv.nl/dspace/handle/1887/3810" for one of the most computationally efficient methods.
     
    Last edited by a moderator: Apr 24, 2017
  6. Feb 10, 2010 #5
    Re: factoring

    an example:
    how can you factor
    x^3 - 2x^2 + 10x + 7 = 0

    It cannot be factored using synthetic division.
     
  7. Feb 10, 2010 #6
    Re: factoring

    Rational Roots Theorem yields x = -1, 1, 7 and -7 as possible roots. x = -1 turns out to be a root. You then divide the polynomial by x + 1.
     
  8. Feb 10, 2010 #7
    Re: factoring

    when i used synthetic division, -1 is not a root of that equation. why should I divide the polynomial by x+1?
     
  9. Feb 10, 2010 #8
    Re: factoring

    Yes, I made a mistake. Well, you can simply solve a qubic equation. First you write it in the form

    y^3 + p y + q

    by substituting

    x = y+2/3

    Then you solve that equation by using the identity:

    (a + b)^3 = a^3 + 3 a^2b + 3 ab^2 + b^3

    You can write this as:

    (a+b)^3 = 3ab(a+b) + a^3 + b^3

    Now the above equation is an identity, which means that it is always valid no matter what you substitute for a and b. Then, if you want to solve an equation of the form:

    y^3 + p y + q = 0

    which we can write as:

    y^3 = -p y - q

    you could try to find numbers a and b such that:

    3 a b = -p

    a^3 + b^3 = -q

    The above identity then implies that y = a + b is a solution.

    Finding a and b is easy, if you take the third power of the first equation, you see that you can write bith equation in the form:

    A B = - (p/3)^3

    A + B = -q

    where A = a^3 and B = b^3

    Fibnding A and B amounts to solving a simple quadratic equation!
     
  10. Feb 11, 2010 #9
    Re: factoring

    Can I know what method you used? and teach me to solve this kind of equations?
     
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