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Homework Help: Factoring cubic polynomial help

  1. Sep 24, 2010 #1
    1. The problem statement, all variables and given/known data
    This is probably an easy question, but using the rational zero theorem I have not found any roots for this cubic polynomial.

    Factor the Following


    2. Relevant equations

    3. The attempt at a solution

    I have used all my knowledge of factoring and have yet to factor this expression. I tried many different values a, such that f(a)=0 using remainder theorem, According to Descartes signs there should be one negative root.
  2. jcsd
  3. Sep 24, 2010 #2


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    Sure you copied it out right? It is fairly close to 6x^3-35x^2-8x+12 which works out quite nice. :devil:
    Last edited: Sep 24, 2010
  4. Sep 24, 2010 #3
    I just learned factoring yesterday, and we were told we could factor all (if i heard correctly) cubic functions, using the factor theorem, remainder theorem, long division, Descartes signs, rational roots theorem as our tools. Yet I am unable to factor this expression. I actually Just made one up. I wanted to know how you would even approach this.
  5. Sep 24, 2010 #4


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    There exists a formula which will tell you the roots of a cubic polynomial and allow you to factor it. However, it's extremely messy, you can see it here:

    To give an example for this function, you can see the solutions on wolfram alpha:

    They aren't particularly nice, and there are only two ways to find them really
    1) Use the cubic root formula
    2) Test various points to see where the polynomial is positive or negative, and narrow down the potential locations for zeros (hence approximating them in the process)
  6. Sep 26, 2010 #5


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    Basically you are asking for almost the main theme of algebra until sometime in the 19th Century.

    How to approach?

    If you are interested enough you can try to what I call hammering them, i.e. use the same approach you have probably learned to factoring quadratics. E.g. in your example you can suppose your cubic a product of three linear factors and the coefficients of x in the brackets are either (6, 1, 1) or (3, 2, 1) while the constant terms are either (12, 1, 1), (6, 2, 1), (3, 4, 1) or (3, 2, 2). I haven't tried all those for your equation but I think none of them work. Quite close is 6x3 - 35x2 - 8x + 12 which factorises as
    ((3x + 2)(2x - 1)(x - 6) I think.

    Also if the roots are integral or rational fractions, they can be obtained by methods you seem to have done.

    What most scientists would do is use a graphic scientific calculator or computer to plot the polynomial and see where it intersected the horizontal axis. Some have root finders or intersection finders and will give you a numerical value for a root. If this is a whole number you can check whether it is an exact solution. Sometimes because of the limitations of digital calculating it may give you instead of say 5.00000000, give you 5.00000001 or 4.999999999 etc. That is almost certainly 5 and you can check that it is.

    But if you ask about how to solve them algebraically... well you first need to be fairly at ease handling imaginary and complex numbers. These are not that difficult and quite fun if anything in maths is.

    Then you remember quadratics are solved by expressing them as the difference of two squares, so for this difference to =0 the squares are equal, and that gives you two solutions as a square has two square roots. In the case of the cubic one method is likewise to try and express it as the difference of two cubes. A cube has three cube roots (two of which may be complex numbers.) To actually get the numbers that make the cubes from the original coefficients you find you have to solve a quadratic equation. Then for the fourth degree you can likewise try to express it as again a difference of two squares but one of them is the square of a quadratic. Similarly to the previous case when you try to find from the original coefficients the numbers or expressions for the things squared, you find you have to solve a cubic equation.

    The formulae are as Office Shredder says rather messy. For which reason, provided you understand complex numbers, it's worth trying to work themselves out yourself, otherwise the textbooks working (and there are lots of books) look like someone else's unmade bed.

    To go further, to solve the fifth degree equation algebraically, it is one of the most celebrated theorems that this is impossible (algebraically being defined as by the arithmetic operations of additions etc. and root extractions). You can only do it algebraically for polynomial of degree higher than 4 if the polynomial is special.

    But you can solve any polynomial numerically. That is, with numerical coefficients like your example there are procedures (also in the books called e.g. Horner's method) for getting a number however close to a root as you want. The electronic calculators are doing something like this, and I think (could be wrong) just home in from both sides of where the polynomial changes sign. Which is not very clever but they make up for that by doing it fast.

    The difference between the algebraic and the numerical approaches is both smaller and bigger than it seems. It may seem the algebra gives you a neat exact formula as contrasted to a messy approximation of the numerical method. But we just said for the cubic and quartic they may be exact but not all that neat. More significantly they are no more exact than the numerical method. E.g. in quadratics you get a square root in the solution. √(b2 - 4ac) may sound like something exact but if that is √5 or √1.1 or √(most numbers) it just stands for an algorithm for an infinite calculation which you stop at some point. So the difference between algebraic and numerical is just the difference between one sort of potentially infinite approximation procedure and another not even very different sort! And BTW in scientific applications I think the algebraic solutions of the cubic and quartic are rarely used compared to the numerical, though I did once meet a chap who had used the quartic one for some aerodynamic problem.

    On the other hand the theory I alluded to was revealing of mathematical structure. They spent centuries not understanding why they couldn't find an algebraic formula like those for degree 1 to 4 for degree 5 and above. The discovery of why opened new paths in maths. But you see at that point the perspective has shifted from your original one of finding a number satisfying an equation to understanding a mathematical structure.

    I may have ventured some opinions with which not everyone will agree here. :uhh:
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