Solving Cubic Equations: Finding the Roots

[Petkovsky]

Find the roots of the equation
x^3 - 3x^2 - 10x +24 = 0

I personally, have never done cubic equations so can you please explain what should i do here. Should i try with GCF, even though i don't see one yet, or is there a method to do this?

 

[Dick]

You're going to have to factor it. There's a method, thanks to the rational root theorem (look it up, ok?) the only possible rational roots of that equation are the divisors of 24 (plus and minus). And if c is a root, then (x-c) is a factor. Divide that factor out and look for another one.

 

[HallsofIvy]

The "rational root theorem" that Dick referred to says that if the fraction (rational number) m/n is root of a polynomial equation axⁿ+ bx^n-1+...+ yx+ z= 0, with coefficients a, b, c, ..., y, z integers, then n must evenly divide a and m must evenly divide z. Here a= 1 and only 1 and -1 evenly divide that. z= 24 and $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 8$, $\pm 12$, and $\pm 24$ evenly divide that. That means that the only possible rational roots are $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 8$, $\pm 12$, and $\pm 24$. Plug them in and see if any of those satisfy the equation. If one does, then you can divide by x- that root to get a quadratic equation solved by the other 2 roots. Of course it is possible that none of them satisfy the equation: that there are no rational roots. In that case, there is a "cubic formula" but it is much more complicated than the quadratic formula and you really don't want to have to do that!

 

[Petkovsky]

Ok so for the first root, by method of trial and error, i got x = 4. I divided the polynomial with x-4, received a quadratic equation and got x2 = -3 and x3 = 2.
I think it's correct, so thank you :)