Finding Roots of a Quartic Polynomial Equation

1. A Quartic Polynomial Equation in 1 Variable is given as follows
   
   \(Ax^4 + Bx^3 + Cx^2 + Dx + E=0\)   ...(1)
   
   where the Co-efficient of \(x^4\) Cannot be Zero i.e. \(A\neq0\)
2. Any Quartic Polynomial Equation in 1 Variable always has 4 Roots/Solutions that are calculated using the following steps
   1. Calculate the Value of Variables \(P\), \(Q\) and \(R\) from equation (1) as follows
      
      \(P=\frac{8AC\hspace{.1cm}-\hspace{.1cm}3B^2}{8A^2}\),   \(Q=\frac{B^3\hspace{.1cm}-\hspace{.1cm}4ABC\hspace{.1cm}+\hspace{.1cm}8A^2D}{8A^3}\),   \(R=\frac{-3B^4\hspace{.1cm}+\hspace{.1cm}256A^3E\hspace{.1cm}-\hspace{.1cm}64A^2BD+\hspace{.1cm}16AB^2C}{256A^4}\)   ...(2)
      
      If the Values of All \(P\), \(Q\) and \(R\) are Zero (i.e. \(P=Q=R=0\)) then the Quartic Polynomial Equation has 1 Root/Solution Repeated 4 times given as
      
      \(x=-\frac{B}{4A}\)   ...(3)
   2. If the Value of Any of \(P\), \(Q\) or \(R\) are Not Zero, calculate the value of variable \(M\) as Root of following Cubic Polynomial Equation using the Cubic Formula
      
      \(8M^3 + 8PM^2 + (2P^2-8R)M -Q^2=0\)   ...(4)
      
      This Cubic Polynomial Equation is called the Resolvent Cubic Equation of the given Quartic Equation.
      
      Once the value of \(M\) is calculated, the Roots of the Quartic Polynomial Equation are calculated using the Quartic Formula as
      
      \(x=\frac{\pm\hspace{1mm}\sqrt{2M}\hspace{1mm}\pm\hspace{1mm}\sqrt{-2P-2M\hspace{1mm}\pm\hspace{1mm}\frac{\sqrt{2}Q}{\sqrt{M}}}}{2}-\frac{B}{4A}\)   ...(5)
3. Please note that the formula given in equation (5) can give 24 Possible Roots/Solution to the Quartic Equation. However Only 4 of them are Valid Roots. Also, solutions provided by equation (5) May have the Same Root Repeated Multiple times although it actually occurs only once in the Quartic Equation. Hence, Only One of the Roots is found using this formula.
4. After One of Roots is calculated using the Quartic Formula given in equation (5), the other 3 Roots are calculated by finding out the Cubic Factor of the Quartic Polynomial and Calculating its Roots using the Cubic Formula. The Cubic Factor of the Quartic Polynomial can be found out by using the Root calculated using the Quartic Formula as follows
   
   Let \(R_1\) be the Root of the Quartic Polynomial Equation calculated using the Quartic Formula
   
   Let \(Kx^3 + Lx^2 + Mx + N\) be the Cubic Factor of the Quartic Polynomial given in equation (1)
   
   Now,   \((x-R_1)(Kx^3 + Lx^2 + Mx + N)=Ax^4 + Bx^3 + Cx^2 + Dx + E\)
   
   \(\Rightarrow Kx^4 + (L-KR_1)x^3 + (M-LR_1)x^2 + (N-MR_1)x - NR_1=Ax^4 + Bx^3 + Cx^2 + Dx + E\)   ...(6)
   
   Using equation (6) the values of \(K, L, M, N\) can be calculated as follows
   
   \(K=A,\hspace{.8cm}L-KR_1=B\hspace{.5cm}\Rightarrow L=B+KR_1,\hspace{.8cm}M-LR_1=C\hspace{.5cm}\Rightarrow M=C+LR_1,\hspace{.8cm}N-MR_1=D\hspace{.5cm}\Rightarrow N=D+MR_1\)
   
   Once the values of \(K, L, M, N\) i.e. the Coefficients and Constant of the Cubic Factor of the Quartic Polynomial are calculated, the Cubic Formula can be used to calculate the other 3 Roots of the Quartic Polynomial.
5. The following gives the Steps for Derivation of Quartic Formula given in equation (5) and Resolvent Cubic Equation as given in equation (4)
   1. Divide the Quartic Polynomial given in equation (1) with the Co-efficient of \(x^4\) i.e. \(A\) as follows
      
      \(x^4 + \frac{B}{A}x^3 + \frac{C}{A}x^2 + \frac{D}{A}x + \frac{E}{A}=0\)   ...(7)
   2. Let \(x=y-\frac{B}{4A}\). Substitute this value of \(x\) in the equation (3) above and simplify to remove the Cube Term from the Quartic Polynomial as follows
      
      \({(y-\frac{B}{4A})}^4 + \frac{B}{A}{(y-\frac{B}{4A})}^3 + \frac{C}{A}{(y-\frac{B}{4A})}^2 + \frac{D}{A}(y-\frac{B}{4A}) + \frac{E}{A}=0\)
      
      \(\Rightarrow y^4 - \frac{B}{A}y^3 + \frac{B^2}{4A^2}y^2 + \frac{B^2}{8A^2}y^2 - \frac{B^3}{16A^3}y + \frac{B^4}{256A^4} + \frac{B}{A}y^3 - \frac{3B^2}{4A^2}y^2 + \frac{3B^3}{16A^3}y - \frac{B^4}{64A^4} + \frac{C}{A}y^2 - \frac{BC}{2A^2}y + \frac{B^2C}{16A^3} + \frac{D}{A}y - \frac{BD}{4A^2} + \frac{E}{A}=0\)
      
      \(\Rightarrow y^4 + (- \frac{B}{A} + \frac{B}{A}) y^3 + (\frac{B^2}{4A^2} + \frac{B^2}{8A^2}- \frac{3B^2}{4A^2} + \frac{C}{A})y^2 + (-\frac{B^3}{16A^3}+ \frac{3B^3}{16A^3}- \frac{BC}{2A^2}+ \frac{D}{A})y + (\frac{B^4}{256A^4} - \frac{B^4}{64A^4} + \frac{B^2C}{16A^3} - \frac{BD}{4A^2} + \frac{E}{A})=0\)
      
      \(\Rightarrow y^4 + (\frac{8AC\hspace{.1cm}-\hspace{.1cm}3B^2}{8A^2})y^2 + (\frac{B^3\hspace{.1cm}-\hspace{.1cm}4ABC\hspace{.1cm}+\hspace{.1cm}8A^2D}{8A^3})y + \frac{-3B^4\hspace{.1cm}+\hspace{.1cm}256A^3E\hspace{.1cm}-\hspace{.1cm}64A^2BD+\hspace{.1cm}16AB^2C}{256A^4}=0\)   ...(8)
      
      Setting \(P=\frac{8AC\hspace{.1cm}-\hspace{.1cm}3B^2}{8A^2}\),   \(Q=\frac{B^3\hspace{.1cm}-\hspace{.1cm}4ABC\hspace{.1cm}+\hspace{.1cm}8A^2D}{8A^3}\)   and   \(R=\frac{-3B^4\hspace{.1cm}+\hspace{.1cm}256A^3E\hspace{.1cm}-\hspace{.1cm}64A^2BD+\hspace{.1cm}16AB^2C}{256A^4}\) we get
      
      \(y^4 + Py^2 + Qy + R=0\)   ...(9)
      
      Any Quartic Polynomial given in the format as represented in equations (8) and (9) above (without the Cube Term) is called a Depressed Quartic Polynomial
   3. From equation (9) we have
      
      \(y^4 + Py^2 + Qy + R=0\)
      
      \(\Rightarrow y^4 + Py^2 =- Qy - R\)   ...(10)
      
      Now, Adding \(\frac{P^2}{4} + M^2 + PM + 2My^2\) on Both Sides of equation (10) we get
      
      \(y^4 + Py^2 + \frac{P^2}{4} + M^2 + PM + 2My^2 =- Qy - R + \frac{P^2}{4} + M^2 + PM + 2My^2\)
      
      \(\Rightarrow {(y^2 + \frac{P}{2}+ M)}^2=2My^2 - Qy + (\frac{P^2}{4} + M^2 + PM - R)\)   ...(11)
      
      Where M is any arbitrarily chosen variable
      
      Now, if M is chosen such that the Quadratic Equation \(2My^2 - Qy + (\frac{P^2}{4} + M^2 + PM - R)\) given in equation (11) above is a perfect square then as per Quadratic Formula
      
      \({(-Q)}^2- 4(2M) (\frac{P^2}{4} + M^2 + PM - R)=0\)   ...(12)
      
      Simplifying equation (12) we get the Resolvent Cubic Equation of the given Quartic Equation as follows
      
      \(8M^3 + 8PM^2 + (2P^2-8R)M -Q^2=0\)
      
      Also since \(2My^2 - Qy + (\frac{P^2}{4} + M^2 + PM - R)\) is a perfect square, it implies
      
      \(\frac{P^2}{4} + M^2 + PM - R=\frac{Q^2}{8M}\)    ...(13)
      
      Therefore
      
      \(2My^2 - Qy + (\frac{P^2}{4} + M^2 + PM - R)= 2My^2 - Qy + \frac{Q^2}{8M}\)
      
      \(\Rightarrow 2My^2 - Qy + (\frac{P^2}{4} + M^2 + PM - R)= {(\sqrt{2M}y - \frac{Q}{2\sqrt{2M}})}^2\)    ...(14)
      
      Using equations (11) and (14)
      
      \({(y^2 + \frac{P}{2}+ M)}^2={(\sqrt{2M}y - \frac{Q}{2\sqrt{2M}})}^2\)
      
      \(\Rightarrow {(y^2 + \frac{P}{2}+ M)}^2-{(\sqrt{2M}y - \frac{Q}{2\sqrt{2M}})}^2=0\)
      
      \(\Rightarrow (y^2 + \frac{P}{2}+ M + \sqrt{2M}y - \frac{Q}{2\sqrt{2M}}) (y^2 + \frac{P}{2}+ M - \sqrt{2M}y + \frac{Q}{2\sqrt{2M}})=0\)
      
      \(\Rightarrow y=\frac{\pm\hspace{1mm}\sqrt{2M}\hspace{1mm}\pm\hspace{1mm}\sqrt{-2P-2M\hspace{1mm}\pm\hspace{1mm}\frac{\sqrt{2}Q}{\sqrt{M}}}}{2}\)
      
      Since \(x=y-\frac{B}{4A}\), therefore
      
      \(x=\frac{\pm\hspace{1mm}\sqrt{2M}\hspace{1mm}\pm\hspace{1mm}\sqrt{-2P-2M\hspace{1mm}\pm\hspace{1mm}\frac{\sqrt{2}Q}{\sqrt{M}}}}{2}-\frac{B}{4A}\)
      
      which is the Formula for Finding Roots/Solutions of Quartic Polynomial Equations.
6. The 4 Roots/Solutions of the Quartic Polynomial Equation in 1 Variable can be of following 28 types depending on the value of variables \(P\), \(Q\), \(R\) and \(M\) as follows
   1. If the Values of All \(P\), \(Q\) and \(R\) are Zero (i.e. \(P=Q=R=0\)) then All the 4 Roots are Same (either Real or Complex). If Value of Any of \(P\), \(Q\) or \(R\) are Not Zero, then the Roots can be as given in the following.
   2. 2 Distinct Pair of Real Roots.
   3. 3 Same 1 Distinct Real Roots.
   4. 4 Distinct Real Roots.
   5. 2 Distinct Pair of Complex Roots.
   6. 3 Same 1 Distinct Complex Roots.
   7. 4 Distinct Complex Roots.
   8. 2 Same Pair of Conjugate Complex Roots.
   9. 2 Distinct Pair of Conjugate Complex Roots.
   10. 2 Same Complex and 1 Distinct Pair of Conjugate Complex Roots.
   11. 2 Distinct Complex and 1 Distinct Pair of Conjugate Complex Roots.
   12. 3 Same Complex and 1 Conjugate Complex Roots.
   13. 2 Same Complex and 1 Conjugate Complex and 1 Distinct Complex Roots.
   14. 3 Same Real and 1 Complex Roots.
   15. 2 Same Real, 1 Distinct Real and 1 Complex Roots.
   16. 3 Distinct Real and 1 Complex Roots.
   17. 2 Same Real and 2 Same Complex Roots.
   18. 2 Same Real and 2 Distinct Complex Roots.
   19. 2 Same Real and 2 Conjugate Complex Roots.
   20. 2 Distinct Real and 2 Same Complex Roots.
   21. 2 Distinct Real and 2 Distinct Complex Roots.
   22. 2 Distinct Real and 2 Conjugate Complex Roots.
   23. 1 Real and 3 Same Complex Roots.
   24. 1 Real, 2 Same Complex Complex and 1 Distinct Complex Roots.
   25. 1 Real and 3 Distinct Complex Roots.
   26. 1 Real, 1 Distinct Complex and 2 Conjugate Complex Roots.
   27. 1 Real , 2 Same Complex and 1 Conjugate Complex Roots.
   You can use the Quartic Equation Roots Calculator to find the Roots/Solutions of Quartic Polynomial Equations.
   
   You can also download the file cubic-quartic.xlsx to calculate Roots/Solutions of Quartic Polynomial Equations (for Quartic Equation with Real Coefficients Only).