Roots of polynomials of degree 3,4,5 with symbolic solution

Roman Chijner, Jan 10, 2022

★ The general form of the 3rd degree equation (or [b]Cubic[/b]) is: ax³ + bx² + cx + d = 0. Cubics have 3 roots. They are given by: ☛∙z1 = cbrt((((b / a / 3 - b / a) b / a / 3 + c / a) b / a / 3 - d / a) / 2 + 0ί + sqrt(((-(b / a / 3 - b / a) b / a / 3 - c / a) b / a / 3 + d / a)² / 4 + ((b / a / 3 - b / a) b / a / 3 + c / a - ((-b) / a / 3 + b / a - b / a / 3) b / a / 3)³ / 27 + 0ί)) + ((((-b) / a / 3 + b / a) b / a / 3 - c / a + ((-b) / a / 3 + b / a - b / a / 3) b / a / 3) / 3 + 0ί) / cbrt((((b / a / 3 - b / a) b / a / 3 + c / a) b / a / 3 - d / a) / 2 + 0ί + sqrt(((-(b / a / 3 - b / a) b / a / 3 - c / a) b / a / 3 + d / a)² / 4 + ((b / a / 3 - b / a) b / a / 3 + c / a - ((-b) / a / 3 + b / a - b / a / 3) b / a / 3)³ / 27 + 0ί)) + (-b) / a / 3; ☛z2 = cbrt(1 / 54) cbrt((-2 b³ - 27a² d + 9a b c + 3sqrt(3) a sqrt(4a c³ + 27a² d² - b² c² + 4b³ d - 18a b c d + 0ί)) / a³) (ί / 2 sqrt(3) - 1 / 2) + 0ί - 1 / 3 b / a + 1 / 3 ((-c) / a + 1 / 3 b / 3 b / a / a + 1 / 3 b * 2 / 3 b / a / a + 0ί) / (cbrt(1 / 54) cbrt((-2 b³ - 27a² d + 9a b c + 3sqrt(3) a sqrt(4a c³ + 27a² d² - b² c² + 4b³ d - 18a b c d + 0ί)) / a³) (ί / 2 sqrt(3) - 1 / 2) + 0ί); ☛z3 = cbrt(1 / 54) cbrt((-2 b³ - 27a² d + 9a b c + 3sqrt(3) a sqrt(4a c³ + 27a² d² - b² c² + 4b³ d - 18a b c d + 0ί)) / a³) ((-ί) / 2 sqrt(3) - 1 / 2) + 0ί - 1 / 3 b / a + 1 / 3 ((-c) / a + 1 / 3 b / 3 b / a / a + 1 / 3 b * 2 / 3 b / a / a + 0ί) / (cbrt(1 / 54) cbrt((-2 b³ - 27a² d + 9a b c + 3sqrt(3) a sqrt(4a c³ + 27a² d² - b² c² + 4b³ d - 18a b c d + 0ί)) / a³) ((-ί) / 2 sqrt(3) - 1 / 2) + 0ί); ★ The general form of the 4th degree equation (or [b]Quartic[/b]) is: ax⁴ + bx³ + cx² + dx + e = 0. Quartics have 4 roots. They are given by: ☛z1 = (-b) / (4a) - 1 / 2 sqrt(0ί + b² / (4a²) - 2c / (3a) + 2^(1 / 3) (c² - 3b d + 12a e) / (3a (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) + 1 / (3 * 2^(1 / 3) a) (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) - 1 / 2 sqrt(0ί + b² / (2a²) - 4c / (3a) - 2^(1 / 3) (c² - 3b d + 12a e) / (3a (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) - 1 / (3 * 2^(1 / 3) a) (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3) - ((-b³) / a³ + 4b c / a² - 8d / a) / (4sqrt(0ί + b² / (4a²) - 2c / (3a) + 2^(1 / 3) (c² - 3b d + 12a e) / (3a (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) + 1 / (3 * 2^(1 / 3) a) (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)))); ☛z2 = (-b) / (4a) - 1 / 2 sqrt(0ί + b² / (4a²) - 2c / (3a) + 2^(1 / 3) (c² - 3b d + 12a e) / (3a (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) + 1 / (3 * 2^(1 / 3) a) (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) + 1 / 2 sqrt(0ί + b² / (2a²) - 4c / (3a) - 2^(1 / 3) (c² - 3b d + 12a e) / (3a (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) - 1 / (3 * 2^(1 / 3) a) (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3) - ((-b³) / a³ + 4b c / a² - 8d / a) / (4sqrt(0ί + b² / (4a²) - 2c / (3a) + 2^(1 / 3) (c² - 3b d + 12a e) / (3a (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) + 1 / (3 * 2^(1 / 3) a) (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)))); ☛z3 = (-b) / (4a) + 1 / 2 sqrt(0ί + b² / (4a²) - 2c / (3a) + 2^(1 / 3) (c² - 3b d + 12a e) / (3a (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) + 1 / (3 * 2^(1 / 3) a) (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) - 1 / 2 sqrt(0ί + b² / (2a²) - 4c / (3a) - 2^(1 / 3) (c² - 3b d + 12a e) / (3a (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) - 1 / (3 * 2^(1 / 3) a) (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3) + ((-b³) / a³ + 4b c / a² - 8d / a) / (4sqrt(0ί + b² / (4a²) - 2c / (3a) + 2^(1 / 3) (c² - 3b d + 12a e) / (3a (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) + 1 / (3 * 2^(1 / 3) a) (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)))); ☛z4 = (-b) / (4a) + 1 / 2 sqrt(0ί + b² / (4a²) - 2c / (3a) + 2^(1 / 3) (c² - 3b d + 12a e) / (3a (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) + 1 / (3 * 2^(1 / 3) a) (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) + 1 / 2 sqrt(0ί + b² / (2a²) - 4c / (3a) - 2^(1 / 3) (c² - 3b d + 12a e) / (3a (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) - 1 / (3 * 2^(1 / 3) a) (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3) + ((-b³) / a³ + 4b c / a² - 8d / a) / (4sqrt(0ί + b² / (4a²) - 2c / (3a) + 2^(1 / 3) (c² - 3b d + 12a e) / (3a (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)) + 1 / (3 * 2^(1 / 3) a) (2c³ - 9b c d + 27a d² + 27b² e - 72a c e + sqrt(0ί - 4(c² - 3b d + 12a e)³ + (2c³ - 9b c d + 27a d² + 27b² e - 72a c e)²))^(1 / 3)))). ★ The general form of the 5th degree equation (or [b]Quintic[/b] ) is: a₅*x⁵+a₄*x⁴+a₃*x³+a₂*x²+a₁x+a₀= 0. Quintics have 5 roots. Because one of the roots of the quintic polynomial is a Real number, let's set it explicitly with a slider: x0. Thus the coefficient a₀ of this fifth degree polynomial is the function p(x0). In the applet, you can use the sliders and corresponding buttons to approximate the desired value of a₀. Using the Ruffini rule, dividing the original polynomial by the binary one, we reduce the order of the polynomial to 4, the solution of which is known in symbolic formulas.


  • 1. Exploring a Conformal Mapping.
  • 2. Cubic polynomial roots using symbolic solutions
  • 3. Solving general cubic in CAS
  • 4. Roots of a cubic polynomial with complex coefficients computed using symbolic solutions
  • 5. Roots of a Cubic polynomial with complex coefficients computed using symbolic solutions
  • 6. Quartic polynomial roots using symbolic solutions
  • 7. Roots of a Quartic polynomial with complex coefficients computed using symbolic solutions
  • 8. Roots of Quintic polynomial computed using symbolic solutions. 1
  • 9. Roots, stationary and inflection points of Quintic polynomial computed using symbolic solutions. 2
  • 10. Roots, stationary and inflection points of Quintic polynomial computed using symbolic solutions. 3
  • 11. Graphical interpretation and visualizing the complex roots of polynomial functions with Real Coefficients.
  • 12. Graphical interpretation and visualizing the roots of complex function(Cubic): f(z)= a z³+ b z² + c z + d . Coefficients a, b, c and d are Complex numbers.
  • 13. Graphical interpretation and visualizing the complex roots of a function(Quartic): f(z)= a z⁴+ b z³ + c z²+ d z+e. Coefficients a, b, c, d and e are Complex numbers.
  • 14. Demo: Determine quintic by 3 stationary points
  • 15. Copy of Determine quintic by 3 stationary points
  • 16. Finding the fourth local extreme point and inflection points of a quintic polynomial constructed by three stationary points using symbolic formulas.
  • 17. Fórmula Cúbica para Polinomios de Grado Tres - Problema de Trigonometría de Dos Ángulos Congruentes - Tangente de Doble Ángulo
  • 18. Formel von Cardano (CAS)

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Exploring a Conformal Mapping.

This applet for the function f(z)=1/z considers two cases of conformal mapping. Here you can explore [br] - the origin of its conformal mapping. It is a mapping for a rectangular grid.[br] - case of a movable regular n-gon and its image under a conformal mapping f.
[b]A rectangular grid and its image under a conformal map f(z)=1/z. [/b]
– Roman Chijner

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