Solving the Cubic and Quartic

Home , Cube root, Cubic equation, Quadratic equation, Quadratic formula, Quartic equation

SOLVING THE CUBIC AND QUARTIC

AARON LANDESMAN

1. INTRODUCTION Likely you are familiar with how to solve a quadratic equation. Given a quadratic of the form ax2 + bx + c, one can ﬁnd the two roots in terms of radicals as √ −b ± b2 − 4ac . 2a On the other hand, the cubic formula is quite a bit messier. The polynomial x4 + ax3 + bx2 + cx + d has roots

. And the quartic formula is messier still. The polynomial x4 + ax3 + bx2 + cx + d has roots

v v u r q u r q u 3 3 2 2 3 2 2 2 2 3 √ u 3 3 2 2 3 2 2 2 2 3 √ 3 b 1 u 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae b2 2c 3 2 c2 − 3bd + 12ae 1 u 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae b2 4c 3 2 c2 − 3bd + 12ae − b + 4cb − 8d x = − − u √ + − + − u− √ + − − − a3 a2 a 1 u 3 2 r u 3 2 r v r 4a 2 t 3 2a 4a 3a 3 q 2 u 3 2a 2a 3a 3 q u 3 q 2 3 3 2 2 3 2 2 2 2 3 3 2 2 3 2 2 2 2 3 3 2 2 3 2 2 2 3√ 3a 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae u 3a 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae u 2c −9bdc−72aec+27ad +27b e+ (2c −9bdc−72aec+27ad +27b e) −4(c −3bd+12ae) 2 2(c2 −3bd+12ae) u u √ b 2c 4 3 + 2 − 3a + r t t 3 2a 4a q 2 3 3a 3 2c3 −9bdc−72aec+27ad2 +27b2 e+ (2c3 −9bdc−72aec+27ad2 +27b2 e) −4(c2 −3bd+12ae)

v v u r q u r q u 3 3 2 2 3 2 2 2 2 3 √ u 3 3 2 2 3 2 2 2 2 3 √ 3 b 1 u 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae b2 2c 3 2 c2 − 3bd + 12ae 1 u 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae b2 4c 3 2 c2 − 3bd + 12ae − b + 4cb − 8d x = − − u √ + − + + u− √ + − − − a3 a2 a 2 u 3 2 r u 3 2 r v r 4a 2 t 3 2a 4a 3a 3 q 2 3 2 u 3 2a 2a 3a 3 q 2 3 u 3 q 2 3 √ 3 2 2 3 2 2 2 3 2 2 3 2 2 2 2c3 −9bdc−72aec+27ad2 +27b2 e+ 2c3 −9bdc−72aec+27ad2 +27b2 e −4 c2 −3bd+12ae 3 3a 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae u 3a 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae u ( ) ( ) b2 2c 2(c2 −3bd+12ae) u 4u √ + − + r t t 3 3 2a 4a2 3a q 3a 3 2c3 −9bdc−72aec+27ad2 +27b2 e+ (2c3 −9bdc−72aec+27ad2 +27b2 e)2 −4(c2 −3bd+12ae)3

v v u r q u r q u 3 3 2 2 3 2 2 2 2 3 √ u 3 3 2 2 3 2 2 2 2 3 √ 3 b 1 u 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae b2 2c 3 2 c2 − 3bd + 12ae 1 u 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae b2 4c 3 2 c2 − 3bd + 12ae − b + 4cb − 8d x = − + u √ + − + − u− √ + − − + a3 a2 a 3 u 3 2 r u 3 2 r v r 4a 2 t 3 2a 4a 3a 3 q 2 3 2 u 3 2a 2a 3a 3 q 2 3 u 3 q 2 3 √ 3 2 2 3 2 2 2 3 2 2 3 2 2 2 2c3 −9bdc−72aec+27ad2 +27b2 e+ 2c3 −9bdc−72aec+27ad2 +27b2 e −4 c2 −3bd+12ae 3 3a 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae u 3a 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae u ( ) ( ) b2 2c 2(c2 −3bd+12ae) u 4u √ + − + r t t 3 3 2a 4a2 3a q 3a 3 2c3 −9bdc−72aec+27ad2 +27b2 e+ (2c3 −9bdc−72aec+27ad2 +27b2 e)2 −4(c2 −3bd+12ae)3

v v u r q u r q u 3 3 2 2 3 2 2 2 2 3 √ u 3 3 2 2 3 2 2 2 2 3 √ 3 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae 2 3 2 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae 2 3 2 b 4cb 8d b 1 u b 2c 2 c − 3bd + 12ae 1 u b 4c 2 c − 3bd + 12ae − 3 + 2 − a x = − + u √ + − + + u− √ + − − + a a . 4 u 3 2 r u 3 2 r v r 4a 2 t 3 2a 4a 3a 3 q 2 3 2 u 3 2a 2a 3a 3 q 2 3 u 3 q 2 3 √ 3 2 2 3 2 2 2 3 2 2 3 2 2 2 2c3 −9bdc−72aec+27ad2 +27b2 e+ 2c3 −9bdc−72aec+27ad2 +27b2 e −4 c2 −3bd+12ae 3 3a 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae u 3a 2c − 9bdc − 72aec + 27ad + 27b e + 2c − 9bdc − 72aec + 27ad + 27b e − 4 c − 3bd + 12ae u ( ) ( ) b2 2c 2(c2 −3bd+12ae) u 4u √ + − + r t t 3 3 2a 4a2 3a q 3a 3 2c3 −9bdc−72aec+27ad2 +27b2 e+ (2c3 −9bdc−72aec+27ad2 +27b2 e)2 −4(c2 −3bd+12ae)3 Whew! That was a mouthful! No one could possibly be expected to memorize such a monstros- ity. And indeed, no one needs to. We will now explain how you can solve the cubic and quartic equations without the mess. Given a cubic or quartic equation, we will explain how to solve it with pure thought. This should convince you that you could write down the solution in radicals if you wanted to. 1 2 AARON LANDESMAN

2. THE CUBIC EQUATION To start, we explain how one might solve the cubic equation. Suppose we start with an equation of the form x3 + αx2 + βx + γ. We would like to find the roots of this equation. Suppose that over the complex numbers, this factors as (x − a)(x − b)(x − c), and we are looking for values of a, b, and c. Then, we can re-express our above equation as x3 + (−a − b − c)x2 + (ab + bc + ca)x − abc. That is, α = −a − b − c, β = ab + bc + ca, γ = −abc. Observe that α, β, γ are simply the symmetric polynomials in a, b, and c. Let ω be a primitive cube root of unity, i.e., ω3 = 1, ω 6= 1. The key to finding a, b, c in terms of radicals is to consider the quantity y := a + ωb + ω2c. It may seem somewhat unmotivated to consider this quantity, but we will explain in section 4 why one would be naturally let to consider this. For the moment, just bear with us. Let τ be the permutation of the roots sending a 7 b, b 7 c, c 7 a Observe that y satisfies the property that → → → τ(y) := τ(a) + ωτ(b) + ω2τ(c) = b + ωc + ω2a = ω2(a + ωb + ω2c) = ω2 · y.

Therefore, τ(y3) = (ω2)3y3 = y3, and so y3 is ﬁxed by τ. Let σ be the permutation of the roots sending a 7 b, b 7 a, c 7 c. Observe that → → → y6 − (y3 + σ(y3))y3 + σ(y3)y3 = 0, and further that (y3 + σ(y3)) and σ(y3)y3 are both ﬁxed by σ. In particular, these are symmetric polynomials, and hence can be expressed in terms of α, β, γ. Therefore, we can express the above equation as (y3)2 − f(α, β, γ)y3 − g(α, β, γ) = 0.

Therefore, y3 satisfies a quadratic equation, and we can use the usual quadratic formula to solve for y3. Then, but taking a cube root, we obtain y. Completely analogously, we can solve for a + ω2b + ωc SOLVING THE CUBIC AND QUARTIC 3 via radicals in terms of α, β, γ. Since γ = a + b + c, we obtain all three of a + b + c a + ωb + ω2c a + ω2b + ωc in terms of radicals. Subtracting the latter two equations from the first, we obtain (1 − ω)b + (1 − ω2)c (1 − ω2)b + (1 − ω)c. Then, we can solve for b and c in terms of the above two equations. In some more detail, if we subtract (1 − ω) times the second equation from the (1 − ω2) times the first, we get (1 − ω2)((1 − ω)b + (1 − ω2)c) − (1 − ω)((1 − ω2)b + (1 − ω)c) = (1 − ω2)2c − (1 − ω)2 c = (−2ω2 + ω + 2ω − ω2)c = (−3ω2 + 3ω)c

So, dividing this by 3ω − 3ω2 yields c. Then, plugging this into the above equations yields a and b. This gives a solution to the cubic equation.

3. THE QUARTIC EQUATION We now explain how to solve the quartic equation, assuming we know how to solve the cubic equation. As above, suppose we have a quartic equation of the form x4 + αx3 + βx2 + γx + δ Suppose we could hypothetically factor this as (x − a) (x − b)(x − c)(x − d). Then, we can again write α, β, γ, δ as the symmetric polynomials in a, b, c, d. To start, we show that we can find the pairwise products ab + cd, ac + bd, ad + bc in terms of α, β, γ, δ. To find these, the key is to consider the resolvent cubic (x − (ab + cd))(x − (ac + bd)) (x − (ad + bc)) . One can observe that this polynomial is symmetric in a, b, c, d and therefore its coefficients can be expressed in terms of α, β, γ, δ. For a Galois-theoretic motivation for why we would consider this resolvent cubic, see section 4. Now, we can use the cubic formula to solve for ab + cd, ac + bd, ad + bc in terms of radicals. We next explain how to find ab, ac, ad, bc, bd, cd in terms of radicals. Let’s say we want to find ab. Then, we know ab + cd and we also know abcd = δ. Therefore, ab is a root of the equation z2 + (ab + cd)z + abcd. So, we can use the quadratic equation to find ab and cd. Similarly, we can find ac, bd, ad, bc in terms of radicals (where ab, cd, ac, bd can be uniquely determined up to reordering, and we may have to guess and check which of ad and bc is which). So, we may assume we have all pairwise products of roots. But now, if we know ab, ac, ad, since we also know δ = abcd, we can compute ab · ac · ad/δ = a2 in terms of radicals, and hence we can find a in terms of radicals. Similarly, we can find b, c, d in terms of radicals, as we wanted. 4 AARON LANDESMAN

4. A GALOISTHEORETICPERSPECTIVE We now explain how to motivate the above solutions in terms of Galois theory. 4.1. Cubic motivation. For solving the cubic, we may observe that the extension of fields Q(a, b, c) over Q(a + b + c, ab + ac + bc, abc) is a Galois extension with Galois group S3. We want to be able to find the values of a, b, c in terms of those of the symmetric polynomials in a, b, c. We are moti- vated to look for simpler subextensions. From Galois theory, these correspond to the fixed fields of subgroups of S3. If we look for the fixed field K of A3 inside S3, we are looking for elements fixed by the action of the permutation τ := (123). Any such element y not in Q will satisfy the property that y/τ(y) = ω for ω a cube root of unity. Then, by the proof of Hilbert’s theorem 90, we can find such an element y, and explicitly take it to be y := a + ωb + ω2c, as is easily checked to satisfy y/τ(y) = ω. One can check y is a generator of Q(a, b, c) over Q(a + b + c, ab + ac + bc, abc) because it is not fixed by any permutation of a, b, c. Then, we can use that the fixed field of A3 is a quadratic extension to solve for y3, and use that Q(a, b, c) over K is cyclic of degree 3 to solve for y (taking the cube root). Since y generates Q(a, b, c) over Q(a + b + c, ab + ac + bc, abc), we can solve for a, b, c in terms of y. One can ask why A3 was the natural subgroup to choose. Indeed, if we are looking for a canonical subgroup of S3, we can take the commutator [S3, S3] = A3. The other nontrivial proper subgroups are isomorphic to Z/2Z and one cannot be canonically chosen as there are three such which are all conjugate. 4.2. Quartic motivation. Let L = Q(a, b, c, d) and K be the field generated over Q by the symmetric polynomials in a, b, c, d. Then, L/K is Galois with Galois group S4. As in the cubic case, we would like to find some subfield intermediate between K and L so that we can solve for a generator of that subfield in terms of radicals. In this case, we seek a canonical subextension. Indeed, take [S4, S4] = A4. If we were to take the fixed field of this, we would get an A4 extension, which we cannot yet solve with the cubic formula. So, let’s take the commutator again! We then get [A4, A4] = K4, where K4 is the Klein-4 group. The fixed field of K4 can be identified explicitly with M := Q (ab + cd, ac + bd, ad + bc) .

We may observe that M/K has Galois group S3 ' S4/K4. So, we can use the cubic formula to ﬁnd ab + cd, ac + bd, ad + bc in terms of α, β, γ, δ.

Remark 4.1. Another way to realize K4 canonically as a subgroup of S4 is to view S4 as the (non orientation preserving) automorphisms of the cube by identifying automorphisms of a cube with permutations of the four main diagonals, A4 as the orientation preserving automorphisms, and K4 as those automorphisms corresponding to 180 degree rotations about a center of any of the six faces.

The cubic formula is used to ﬁnd ab + cd, ac + bd, ad + bc, which is an S3 extension of K. Then, S4 is a further extension of S3 by K4 ' Z/2 × Z/2, which we can then further solve using the quadratic formula twice.

5. EXERCISES Doing the following exercises is a sure way to understand the above methods, but it is probably harder to solve them than to understand how to solve them. Exercise 5.1 (Only for the most hardcore). Using the above method to solve the cubic, ﬁnd explicit roots to the equation y3 − 4y − 1. Exercise 5.2 (Not for the faint of heart). Using the above method to solve the quartic, ﬁnd explicit roots to the equation x4 + x + 1.