Solving this cubic equation $x^3+15x^2+24x-40=0$ algebraically

Question

I have : $x^3+15x^2+24x-40=0$

When I use $x=u-a/3$ where $a=15$ and I replace it gives :

$u^3-51u+90=0$

Now, my discriminant is inferior to $0$...

How do I find out at least one solution of this cubic equation without using trigonometric methods? I only learned how to solve cubics with algebraical methods...

Thank you!

Answer 1

Notice that $x=1$ is an easily obtainable root of your cubic equation. As Clement points out in the comment, the first root is usually easily obtainable. You can see the constant term of your cubic, it is $-40$, so the product of your roots is $40$. So you can try some simple factors of $40$ like $1,-1,2$ etc.

Factoring it out yields $$(x-1)(x^2+16x+40)=0$$

Solving the quadratic, you get the other two roots.

Answer 2

When in a cubic equation coefficient of the highest power of the variable(here the variable is x)is1 then try out the different factors of the constant term of the equation.So try out the factors of 40 which are 1,2,4,5,8,10,20,40 in both signs that is positive as well as negative factors.By trying+1 you get 1 as a solution.Then divide the equation by x-1.Then you get a quotient which is qudratic which you can solve easily.This method is similar to factor theorem.