How to solve the cubic equation $ x^3+3x -2 = 0$ without using matrices? I am trying to solve $ x^3+3x -2 = 0$ Using the remainder theroem but none of the factors of the constant make the equation equal to $0$. Is there any way I can get the answers without using matrices?
 A: You can use differential calculus to find the roots of the given function.
First lets find the peaks using $dy/dx = 0$
let $f(x) = x^3 + 3x -2$
it seems that $$\frac{dy}{dy} = 3x^2 +3 =  0$$
$$x = \sqrt{-1},\quad -\sqrt{-1}$$
This implies that the function does not have any peaks, this means that
it has 1 real root, and 2 imaginary roots, and also it is strictly increasing function, as the coefficient of the largest degree term is positive.
Now lets find the real root, 
$$f(0) = -2$$
$$f(1) = 2$$
This means it has the root somewhere in between $0$ and $1$.
Now, Using Newton-Raphson method, choose $x = 1$ as root first,
$$x_1 = 1 - \frac{f(1)}{f^{'}(1)}$$
$$x_1 = {2\over3} = 0.66666$$
We can now guarantee that the root is between $0$ and $2/3$
Second iteration gives the root as, $$x_2 = {2\over3} - \frac{f(2/3)}{f^{'}(2/3)}$$
$$x_2 = {70\over117} = 0.59829$$
So the third approximation will only correct the value after 4 decimal places, so its better to stop after 2nd iteration over here, and get the real root as $x_2 = 0.5983$
By this method you can solve not only cubic, biquadratic equations but also some complex equations involving logarithmic or exponential function.
A: I learned a method that predates home PCs but came after 4 function calculators. 
Manipulate to get x=2/((x^2)+3) 
Enter 1 and you get x=1/2. Then enter that result in for x (the x of x^2 not first x) and it will converge damn fast. After 8 iterations I have .59607 . 
The method works, but of course won't tell you of this is a good-looking irrational number. 
Edit - I appreciate the answer using calculus, but the question itself looks like early high school, and considered my 4-function calculator answer clever. 
