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Solving the Cubic and Quartic

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Solving the Cubic and Quartic 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 p -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 p u 3 3 2 2 3 2 2 2 2 3 p 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 p + - + - u- p + - - - 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 3p 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 p 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 p u 3 3 2 2 3 2 2 2 2 3 p 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 p + - + + u- p + - - - 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 p 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 p + - + 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 p u 3 3 2 2 3 2 2 2 2 3 p 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 p + - + - u- p + - - + 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 p 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 p + - + 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 p u 3 3 2 2 3 2 2 2 2 3 p 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 p + - + + u- p + - - + 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 p 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 p + - + 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 ! be a primitive cube root of unity, i.e., !3 = 1, ! 6= 1. The key to finding a, b, c in terms of radicals is to consider the quantity y := a + !b + !2c. 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) + !2τ(c) = b + !c + !2a = !2(a + !b + !2c) = !2 · y. Therefore, τ(y3) = (!2)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 + !2b + !c 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 + !b + !2c a + !2b + !c in terms of radicals. Subtracting the latter two equations from the first, we obtain (1 - !)b + (1 - !2)c (1 - !2)b + (1 - !)c. Then, we can solve for b and c in terms of the above two equations. In some more detail, if we subtract (1 - !) times the second equation from the (1 - !2) times the first, we get (1 - !2)((1 - !)b + (1 - !2)c)-(1 - !)((1 - !2)b + (1 - !)c) = (1 - !2)2c - (1 - !)2 c = (-2!2 + ! + 2! - !2)c = (-3!2 + 3!)c So, dividing this by 3! - 3!2 yields c.
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