Finding root of a polynomial Let's say I've got $ax^3 + bx^2 + cx + d = 0$
I want to find x for a general-case.
Mathematica gave me a very long solution, and even longer for one which has $kx^4$.
Can you please help me reproduce the algorithm used by Mathematica to generate formula of:
$$P(x) = a_0x^0 + a_1x^1 + a_2x^2 + a_3x^3 + \cdots+ a_nx^n$$
i.e., general case of a polynomial.
 A: The trick is transforming the general cubic equation into a new cubic equation with missing $x^2$ term is due to
Nicolo Fontana Tartaglia.  
Apply the substitution
$$ x = y - \frac{b}{3a} $$
then we get 
$$ a\left(y-\frac{b}{3a} \right)^3 + b\left(y-\frac{b}{3a} \right)^2+c\left(y-\frac{b}{3a} \right)+d = 0 $$
which simplifies to
$$ ay^3 + \left( c-\frac{b^2}{3a}\right)y+ \left(d + \frac{2b^3}{27a^2} - \frac{bc}{3a} \right) = 0 $$
This is called a depressed cubic equation, because the square term is eliminated.  It is much easier to use this and then
find the roots. (back substitute to get the roots in terms of $x$)
For example $$2x^3-18x^2+46x-30=0$$
Substitute $ x=y+3$ and simplify this cubic equation to
$2y^3-8y=0 \Rightarrow y=0,2,-2$ which then gives the roots as
$x=1,3, $ and $5$.
A: Here's a (standard) method to treat the cubic case, though it hides some of what is going on, which is basically Galois Theory. You might as well suppose that $a \neq 0$ (otherwise you would be dealing with a quadratic at worse). Hence you can divide through and assume that $a = 1.$ So let's work instead with the cubic $x^3 + ax^2 + bx + c.$ Now set $y = x + \frac{a}{3}.$ Then $x^3 + ax^2 + bx + c$ has the form $y^3 + ey + f,$ and if we can solve for $y,$ we can easily recover $x$ (the constants $e$ and $f$ can be calculated routinely in terms of $a,b$ and $c,$ but we omit the rather tedious details. Hence if we can solve equations of the form $x^3 + ax + b =0,$ we can deal with the general cubic, and we may as well suppose that $b \neq 0,$ for the case $b =0$ is easy.  To solve the equation $x^3 + ax + b = 0$ when $b \neq 0$,  set $x = u+v.$ The equation may then be written as
$(u^3 + v^3) + b + (u+v)(3uv+a) = 0.$ We can try to solve separately, but simultaneously, $(u+v)(3uv+a) = 0$ and $u^3 +v^3 = -b.$ Note that $u +v \neq 0$ since $b \neq 0,$ so we need $uv = -\frac{a}{3}$ and $u^3 + v^3 = -b.$ Hence $(t -u^3)(t-v^3) = t^2 +bt -\frac{a^3}{27}$, and two solutions do have a simultaneous solution, with $u^3$ and $v^3$ being the roots of a quadratic.
A: Although there are formulas for 3rd and 4th  degree polynomials, it can be proven that there are no formulas for 5th degree polynomials or higher. This statement is known as the Abel-Ruffini Theorem. A proof of this statement can be found on Wikipedia (https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem).
