Raymond B. answered • 04/18/20
Math, microeconomics or criminal justice
It's not easy to solve a cubic equation.
If there is a simple rational solution, it's usually an integer or a simple fraction composed of factors of the constant term and the coefficient of the 3 degree term, 2/3, 1/3, 1/1, 2/1, or any with a negative sign
You try them, plug them in and see if f(x) equals zero. IF not then you can't factor the cubic. If you could get a solution, then you can divide and get a quadratic, use the quadratic formula to get 2 more solutions, which may be rational, irrational or imaginary.
Another strategy is plot a few points. Take the derivative to find the local extrema
The derivative is 6x^2 + 6x -6. Set it equal to zero. Then solve for x.
Divide by 6, to get x^2 +x -1 = 0. Use the quadratic formula to get x=about .6 and -1.6.
Those respectively, are local minimum and local maximum. Both are positive when plugged
into the original cubic equation. The only possible rational root is the real root, which is
to the left of the local max, about -2.8, which when plugged into the cubic equation yields
nearly zero.
Make a factor with -2.8, which is x+2.8. Divide it into the cubic equation to get
approximately 2x^2 - 2.6x + 1.28. Solve that for x with the quadratic equation, to
get 0.325 + or - .47i
That's 2 imaginary roots. Neither is rational.
The approximately -2.8 root may or may not be rational.
Maybe someone has a better solution. But -2.8 give or take .1 is the only possible rational root.
There are complicated methods for solving cubic equations, such as finding the discriminates for a quadratic and a cubic, combine them in another complicated formula. But it gets complicated as well as the calculations.
25/9 is close to a solution, but not quite. Since no simple fraction works, odds are there's only an irrational solution. The complicated method involves square roots and cube roots which generally end up in irrational solutions.
