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Factoring Help

  1. Oct 4, 2005 #1
    How can we factor a question which is too the power of 4 or quartic. for example ax^4+bx^3+cx^2+dx+e, but a has a grreater value than 1.
  2. jcsd
  3. Oct 4, 2005 #2
    I'm not sure, but I think that fourth order equations are solvable with some weird equation like third order. In practice, you might start with the rational zeros test. What you do is: take all possible factors of 'e' and divide by all possible factors of 'a'. Then test these to see if they are zeros (if q is a zero, x-q is a factor). Using a procedure called synthetic division lets you use the factors that you find to reduce the order of the equation. When the equation has finally been reduced down to a quadratic, then you can use the quadratic formula. Alternately, there are 'numerical' methods that can find approximations of the zeros, (i.e. Newtons Method). There are a few other tricks. For example. If it looks like ax^4+bx^2+c, you can just use the quadratic formula to solve for x^2, and then take the square root of your answers.
    In general factoring higher order polynomials is a tricky business. Hope I answered your question :smile:
  4. Oct 5, 2005 #3


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    Science Advisor

    Theoretically, any polynomial equation can be factored completely into linear factors if you use complex numbers, linear and quadratic factors if you stay in real numbers.

    Yes, there exist a "quartic" formula. Here is a website with a (comparatively) simple explanation: http://web.usna.navy.mil/~wdj/book/node95.html [Broken]

    There is, however, no general method for factoring except by solving the equation and then using the solutions to get the factors: If a, b, c are solutions, the we can factor as p(x-a)(x-b)(x-c).
    Last edited by a moderator: May 2, 2017
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