Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was trying to solve a cubic equation which is :

$ -\lambda³ -\lambda² + 10 \lambda - 8 = 0$

I googled about it and I found the Rational Root theorem which is takes time to do it, but I found that some people are solving cubic equation by factoring, like this equation :

x3+4x2+x-6 = 0 it will be (x-1)(x+2)(x+3)=0

I tried to do the same with my equation but it's hard to find what is the common factorial.

Is there any way to determine what is the common factorial as in the example I gave you (x-1)(x+2)(x+3)=0

share|cite|improve this question
"Factorials", that is a bit misleading. – Sawarnik Nov 10 '13 at 16:19

In this case the rational root theorem tells you that the only possible rational roots are $\pm 1, \pm 2, \pm 4, \pm 8$ and you can just plug them into your polynomial and see if any are roots. It turns out $1$ works. So $\lambda -1$ is a factor and you can divide it out, getting $(\lambda - 1)(-\lambda^2-2\lambda +8)=0$ Now you have a quadratic which you can use the quadratic formula on or complete the square, getting $-(\lambda-1)(\lambda+4)(\lambda-2)=0$. In this case, three of the eight possibilities work.

share|cite|improve this answer

$-(1)^3-(1)^2+10(1)-8=0$. So we know $(\lambda-1)$ is a factor. You'll be left with a quadratic after long division. The factoring method will rely on guessing the first root so in most cases you'll have an easy polynomial. $\pm 1, \pm 2, \pm 3$ are most common guesses.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.