# Generalizations of the quadratic formula [duplicate]

The quadratic formula can be used to find the roots of any quadratic polynomial of the form $ax^2 + bx + c$:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The derivation is simple enough and uses a technique called completing the square.

Is there a formula to solve cubic equations of the form:

$$ax^3 + bx^2 + cx + d$$

Or a general formula to solve an $n$th degree polynomial?

EDIT: According to the answers below, and some mathematicians, such a formula isn't possible. But why is that? I mean, consider an $n$th degree polynomial $p(x)$ with roots $r_1, r_2, \dots, r_n$:

$$p(x) = ax^n - \left(\sum r_i\right)ax^{n-1} + \left(\sum_{i, j} r_ir_j\right)ax^{n-2} + \dots+(-1)^n(r_1r_2\dots r_n)$$

We have $n$ equations with $n$ unknowns, surely this can be solved using a general algorithm?

## marked as duplicate by user61527, dfeuer, Hanul Jeon, Dominic Michaelis, Jyrki LahtonenNov 14 '13 at 6:56

• I suspect this is a duplicate, but I can't find it anywhere, so maybe not. There are formulas to find roots of polynomials of degree $3$ and $4$, which can be found on the Wikipedia pages for "cubic function" and "quartic function"; the quartic one is particularly nasty, but even the cubic one is fairly unwieldy in practice. For $n\geq 5$, general polynomials of degree $n$ are not solvable in radicals, so no formula exists. (The proof of this uses Galois theory). – mdp Nov 7 '13 at 16:06
• This may be the only question ever tagged both algebra-precalculus and Galois theory. Gerard, you do indeed have $n$ factors. The trouble is that they can't necessarily be expressed in any nice way. To put it more precisely, the set of complex algebraic numbers (roots of some polynomial with integer coefficients) is a proper superset of the numbers expressible using just sums, products, ratios, powers, and $k$th roots. – dfeuer Nov 14 '13 at 3:35
Yes, there is for $n=3$ and $n=4$, but not for $n\geq 5$. This is due to results in Galois Theory.