Complex numbers and polynomials - University College Cork

q = −3uv, r = u 3 + v 3 for u, v. It’s easy to see that u 3 , v 3 must be the roots ofthe quadratic equation.Hence,z 2 − rz − q327 = 0.√u 3 , v 3 = r ± r 2 + 4q327.2So, let u be a cube root of one of these and let v = −q/(3u). Then −(u + v)is a root, √ possibly complex, of x 3 + qx + r. Having selected one cube root u of(r + r 2 + 4q3 )/2, say, and setting v = −q/3u, the remaining roots can then27be found by solving x 2 −√(u + v)x + u 2 + v 2 − uv = 0, or by figuring out theother cube roots of (r + r 2 + 4q3 )/2. 27Notice that u 3 is real iff 27r 2 + 4q 3 ≥ 0. So, if this condition is satisfied,then −(u + v) is a real root—possibly the only real root, as the followingexample shows. (In fact, it’s not too hard to see that if 27r 2 + 4q 3 > 0holds, then −(u + v) is the only real root of the cubic. What may happen if27r 2 + 4q 3 = 0?)Example 7. Solve x 3 + 3x + 1 = 0.Solution. For this, uv = −1, and u 3 , v 3 are roots of z 2 − z − 1 = 0. So,u 3 , v 3 = 1 ± √ 5.2√Say u = 3 1+ √ √5. Then (?) v = 3 1− √ 5, and so2 2√1 + √ √5 1 − √ 5− 3 − 32 2is a real root of x 3 + 3x + 1. The remaining roots are given by(u + v) ± √ (u + v) 2 − 4(u 2 + v 2 − uv)2and are complex numbers.= (u + v) ± √ 3i(u − v),24. The last remark raises an interesting question: what conditions on the coeffsq, r guarantee that all three roots of x 3 +qx+r are real? To see what’s involved,let a, b, c be the real roots of this cubic. Then a + b + c = 0, ab + bc + ca =q, abc = −r. Hence r 2 = a 2 b 2 c 2 . Since the geometric mean of positive numbersdoesn’t exceed their arithmetic mean, we see that3√r2 ≤ a2 + b 2 + c 23= (a + b + c)2 − 2(ab + bc + ca)327= −2q3 .