Cubic and Quartic Formulas

All algebra students eventually come into contact with the quadratic formula. It provides a nice way to solve quadratic equations. The obvious question is whether or not similar formulas exist for higher degree polynomials. For degrees 5 and above, it is a non-trivial fact from abstract algebra that no such formula exists. However, for degrees 3 and 4 there are similar solutions.

Cubic Formula

Consider x³ + Ax² + Bx + C. Let X = x + A/3. Then

    where Q =  and R = .

Suppose a³ + Qa + R = 0. Write a = b + c where b and c are to be determined. Then a³ = (b + c)³ = b³ + c³ + 3(cb² + bc²) = b³ + c³ + 3abc.  Then b³ + c³ + (3bc + Q)a + R = 0. (*) Also suppose bc = -Q/3. (Don't worry about why, since it works.) Then the middle term in (*) vanishes.

Therefore (*) becomes b³ + c³ = -R and b³c³ = -Q³/27. Thus, from substitution we have

,

or

 By the quadratic formula we have

 and, since b³ + c³ = -R,

Thus

where the cube roots are chosen so the product bc = -Q/3. This is a root of the altered cubic equation so x = a - A/3 Once that root is divided out you have a quadratic which can be solved easily.

Quartic Formula

For the quartic x⁴ + Ax³ + Bx² + Cx + D, let x = X - A/4. This gives

where

X⁴ + QX² + RX + S = (X² + KX + L)(X² - KX + M) where finding K, L and M reduces us to two quadratics.

Expanding the right hand side and equating coefficients of like powers of x gives L + M - K² = Q, K(M - L) = R and LM = S.

Thus, from the first two of these we get 2M = K² + Q + R/K and 2L = K² + Q - R/K. Substituting these into the third equation gives (K³ + QK + R)(K³ + QK - R) = 4SK² which becomes K⁶ + 2QK⁴ + (Q² - 4S)K² - R² = 0 which is a cubic in K² and can thus be solved using the technique given above. With K known, we can find L and M, giving us the two quadratics which we can solve.

These formulas come from the book An Introduction to the Theory of Groups by Joseph Rotman.