Solving the cubic equation $ -\lambda³ -\lambda² + 10 \lambda - 8 = 0$

Question

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:$$x^3+4x^2+x-6 = 0 ;$$it will be$$(x-1)(x+2)(x+3)=0 .$$

I tried to do the same with my equation, but it is 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, $(x-1)(x+2)(x+3)=0$?

Answer 1

$-(1)^3-(1)^2+10(1)-8=0$. 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.

Answer 2

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.