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Factoring a 3rd degree poly to get a start on partial fractions

  1. Jun 12, 2013 #1
    Hello all,

    I'm working through old exams for an electrical subject (no solutions given) and I think I've gone wrong somewhere and been left with something I'd like to learn how to work with anyway:


    [itex]\frac{50}{2s+3+\frac{2}{s}+\frac{1}{s^2}}\times \frac{s^2}{s^2}[/itex]



    From the last step I can do partial fractions and inverse Laplace etc...

    But I got that factorisation via matlab, I can see it obviously works, and given a guess at a factor of the denominator of the 2nd to last line I can see that '+1' would be involved so a next guess might be '(s+1)' and then I'd see if I get zero remainder with polynomial long division, and I'd win in this case, but it was just a lucky guess.

    Ok, time for the question: what is the direct method for doing this ?

    I left the earlier working up the case we could hijack the math earlier in the process.

  2. jcsd
  3. Jun 12, 2013 #2


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    There is a formula for solutions to the cubic polynomial but nobody uses it in practice. Instead they use the fact that if there is a rational root to [itex] ax^3 + bx^2 + cx + d [/itex] where a,b,c,d are all integers, then the root has to be of the form +/-m/n where m and n are integers, coprime, and m divides d, and n divides a (this is called the rational root theorem)

    For your cubic, the only possible rational roots are 1,-1, 1/2 and -1/2 from this. So you can just try them all and see that -1 is a root and factor out an s+1
  4. Jun 13, 2013 #3


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    The thing to have seen was that this is just the standard "difference of two squares"
    (a2 - b2) = (a - b)(a + b)

    Remember now? :smile:
  5. Jun 14, 2013 #4

    I've missed a lot of basic stuff at university (sort of jumped ahead via some credit I perhaps shouldn't have got) and school was a long time ago so a lot of these rules of thumbs and tricks go way over my head...

    Although here I am doing a masters in engineering - you might say 'lol' - if you were that way inclined :tongue2:
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