Finding Roots of a Cubic Polynomial Equation

1. A Cubic Polynomial Equation in 1 Variable is given as follows
   
   \(Ax^3 + Bx^2 + Cx + D=0\)   ...(1)
   
   where the Co-efficient of \(x^3\) Cannot be Zero i.e. \(A\neq0\)
2. Any Cubic Polynomial Equation in 1 Variable always has 3 Roots/Solutions that are calculated using the following steps
   1. Calculate the Value of Variables \(P\) and \(Q\) from equation (1) as follows
      
      \(P=\frac{3AC\hspace{.1cm}-\hspace{.1cm}B^2}{3A^2}\),   \(Q=\frac{2B^3\hspace{.1cm}-\hspace{.1cm}9ABC\hspace{.1cm}+\hspace{.1cm}27A^2D}{27A^3}\)   ...(2)
      
      If the Value of Both \(P\) and \(Q\) are Zero (i.e. \(P=Q=0\)) then the Cubic Polynomial has 1 Root/Solution Repeated 3 times given as
      
      \(x=-\frac{B}{3A}\)   ...(3)
   2. If the Value of either \(P\) or \(Q\) are Not Zero, then the Roots of the Cubic Polynomial Equation are calculated using the following Cubic Formula
      
      \(x=\sqrt[3]{\frac{-Q}{2} + \sqrt{D_c}} + \sqrt[3]{\frac{-Q}{2} - \sqrt{D_c}} - \frac{B}{3A}\)   ...(4)
      
      where \(D_c\) is called the Discriminant of the Cubic Polynomial and is calculated as
      
      \(D_c= {(\frac{Q}{2})}^2 + {(\frac{P}{3})}^3 \)   ...(5)
3. Please note that the formula given in equation (4) can give 9 Possible Roots/Solution to the Cubic Equation. However Only 3 of them are Valid Roots. Also, solutions provided by equation (4) May have the Same Root Repeated Multiple times although it actually occurs only once in the Cubic Equation. Hence, Only One of the Roots is found using this formula.
4. After One of Roots is calculated using the Cubic Formula given in equation (4), the other 2 Roots are calculated by finding out the Quadratic Factor of the Cubic Polynomial and Calculating its Roots using the Quadratic Formula. The Quadratic Factor of the Cubic Polynomial can be found out by using the Root calculated using the Cubic Formula as follows
   
   Let \(R_1\) be the Root of the Cubic Polynomial Equation calculated using the Cubic Formula
   
   Let \(Kx^2 + Lx + M\) be the Quadratic Factor of the Cubic Polynomial given in equation (1)
   
   Now,   \((x-R_1)(Kx^2 + Lx + M)=Ax^3 + Bx^2 + Cx + D\)
   
   \(\Rightarrow Kx^3 + (L-KR_1)x^2 + (M-LR_1)x - MR_1=Ax^3 + Bx^2 + Cx + D\)   ...(6)
   
   Using equation (6) the values of \(K, L, M\) can be calculated as follows
   
   \(K=A,\hspace{.8cm}L-KR_1=B\hspace{.5cm}\Rightarrow L=B+KR_1\hspace{.5cm}\Rightarrow L=B+AR_1,\hspace{.8cm}M-LR_1=C\hspace{.5cm}\Rightarrow M=C+LR_1\)
   
   Once the values of \(K, L, M\) i.e. the Coefficients and Constant of the Quaratic Factor of the Cubic Polynomial are calculated, the Quadratic Formula can be used to calculate the other 2 Roots of the Cubic Polynomial as follows
   
   \(x=\frac{-L\hspace{.1cm}\pm\hspace{.1cm}\sqrt{L^2-4KM}}{2K}\)...(7)
5. The following gives the Steps for Derivation of Cubic Formula given in equation (4)
   1. Divide the Cubic Polynomial given in equation (1) with the Co-efficient of \(x^3\) i.e. \(A\) as follows
      
      \(x^3 + \frac{B}{A}x^2 + \frac{C}{A}x + \frac{D}{A}=0\)   ...(8)
   2. Let \(x=y-\frac{B}{3A}\). Substitute this value of \(x\) in the equation (8) above and simplify to remove the Square Term from the Cubic Polynomial as follows
      
      \({(y-\frac{B}{3A})}^3 + \frac{B}{A}{(y-\frac{B}{3A})}^2 + \frac{C}{A}(y-\frac{B}{3A}) + \frac{D}{A}=0\)
      
      \(\Rightarrow y^3 -\frac{B}{A}y^2 + \frac{B^2}{3A^2}y - \frac{B^3}{27A^3} + \frac{B}{A}y^2 -\frac{2B^2}{3A^2}y + \frac{B^3}{9A^3} + \frac{C}{A}y - \frac{BC}{3A^2} + \frac{D}{A} =0\)
      
      \(\Rightarrow y^3 + \frac{C}{A}y - \frac{B^2}{3A^2}y - \frac{B^3}{27A^3} + \frac{B^3}{9A^3} - \frac{BC}{3A^2} + \frac{D}{A} =0\)
      
      \(\Rightarrow y^3 + (\frac{C}{A} - \frac{B^2}{3A^2})y + (\frac{2B^3}{27A^3} - \frac{BC}{3A^2} + \frac{D}{A}) =0\)
      
      \(\Rightarrow y^3 + \frac{3AC\hspace{.1cm}-\hspace{.1cm}B^2}{3A^2}y + \frac{2B^3\hspace{.1cm}-\hspace{.1cm}9ABC\hspace{.1cm}+\hspace{.1cm}27A^2D}{27A^3} =0\)   ...(9)
      
      Setting \(P=\frac{3AC\hspace{.1cm}-\hspace{.1cm}B^2}{3A^2}\) and \(Q=\frac{2B^3\hspace{.1cm}-\hspace{.1cm}9ABC\hspace{.1cm}+\hspace{.1cm}27A^2D}{27A^3}\) we get
      
      \(y^3 + Py + Q =0\)   ...(10)
      
      Any Cubic Polynomial given in the format as represented in equations (9) and (10) above (without the Square Term) is called a Depressed Cubic Polynomial
   3. Now, we know that for any 2 variables \(u\) and \(v\)
      
      \({(u+v)}^3=u^3 + 3u^2v + 3uv^2 + v^3\)
      
      \(\Rightarrow {(u+v)}^3 -3uv(u+v) - (u^3 + v^3)=0\)   ...(11)
      
      Now, Setting \(y=u+v\) and comparing equations (10) and (11) we get
      
      \(P=-3uv\hspace{.5cm}\Rightarrow uv=-\frac{P}{3}\)   ...(12)
      
      \(Q=- (u^3 + v^3)\hspace{.5cm}\Rightarrow u^3 + v^3+Q=0\)   ...(13)
   4. Multiplying equation (13) with \(u^3\) we get
      
      \({(u^3)}^2 +Qu^3 + {(uv)}^3 =0\)   ...(14)
      
      Putting the value of \(uv\) from equation (12) in equation (14) we get
      
      \({(u^3)}^2 +Qu^3 - {(\frac{P}{3})}^3 =0\)   ...(15)
      
      Since the equation (15) is Quadratic in terms of \(u^3\), the value of \(u^3\) (and hence the value of \(u\)) can be calculated using Quadratic Formula as follows
      
      \(u^3= \frac{-Q \pm \sqrt{Q^2 + 4{(\frac{P}{3})}^3}}{2}=\frac{-Q}{2} \pm \sqrt{ {(\frac{Q}{2})}^2 + {(\frac{P}{3})}^3 }\)   ...(16)
      
      The term \({(\frac{Q}{2})}^2 + {(\frac{P}{3})}^3 \) given in the equation (16) above is called the Discriminant of the Cubic Polynomial Equation and is denoted by \(D_c\). Therefore,
      
      \(u = \sqrt[3]{\frac{-Q}{2} \pm \sqrt{ {(\frac{Q}{2})}^2 + {(\frac{P}{3})}^3 }}= \sqrt[3]{\frac{-Q}{2} \pm \sqrt{ D_c }} \)   ...(17)
      
      Similarly, if we multiply equation (13) with \(v^3\) and solving for value of \(v\) we get
      
      \(v = \sqrt[3]{\frac{-Q}{2} \pm \sqrt{ {(\frac{Q}{2})}^2 + {(\frac{P}{3})}^3 }}= \sqrt[3]{\frac{-Q}{2} \pm \sqrt{ D_c }} \)   ...(18)
      
      Please note that both variables \(u\) and \(v\) have identical set of values. Also, in case of Repeated Roots for \(u^3\) / \(v^3\), \(D_c=0\) and hence we have
      
      \(u^3=v^3=\frac{-Q}{2}\hspace{5mm}\Rightarrow u=v = \sqrt[3]{\frac{-Q}{2}}\)   ...(19)
   5. Now, Since \(y=u+v\), substituting the values of \(u\) and \(v\) from equations (17) and (18) we get
      
      \(y=\sqrt[3]{\frac{-Q}{2} \pm \sqrt{D_c}} + \sqrt[3]{\frac{-Q}{2} \pm \sqrt{D_c}}\)   ...(18)
   6. The value of \(x\) for the Cubic Polynomial Equation given in equation (1) and (8) can be found out by Subtracting \(\frac{B}{3A}\) from the Value of \(y\) calculated in equation (18) as follows
      
      \(x=\sqrt[3]{\frac{-Q}{2} \pm \sqrt{D_c}} + \sqrt[3]{\frac{-Q}{2} \pm \sqrt{D_c}} - \frac{B}{3A} \)   ...(20)
      
      However, the 2 Square Roots of the Discriminant \(D_c\) Only Differ in Sign. Therefore
      
      \(x=\sqrt[3]{\frac{-Q}{2} + \sqrt{D_c}} + \sqrt[3]{\frac{-Q}{2} - \sqrt{D_c}} - \frac{B}{3A} \)   ...(21)
      
      which is the Formula for Finding Roots/Solutions of Cubic Polynomial Equations.
6. The 3 Roots/Solutions of the Cubic Polynomial Equation in 1 Variable can be of following 13 types depending on the value of variables \(P\), \(Q\) and \(D_c\) as follows
   1. If the Value of Both \(P\) and \(Q\) are Zero (i.e. \(P=Q=0\)) then All the 3 Roots are Same (either Real or Complex). If the Value of either \(P\) or \(Q\) are Not Zero, then the Value of the Discriminant \(D_c\) determines the Type of Roots as given in the following steps.
   2. If \(D_c\) is Real and \(D_c<0\) then All the 3 Roots are Real and Distinct.
   3. If \(D_c\) is Real and \(D_c>0\) then One Root is Real and 2 Roots are Conjugate Complex.
   4. If \(D_c=0\) then Atleast 2 Roots are Same. This can have the following subtypes
      1. 2 Same 1 Distinct Real Roots
      2. 2 Same 1 Distinct Complex Roots
      3. 2 Same 1 Conjugate Complex Roots
      4. 2 Same Real 1 Complex Roots
      5. 2 Same Complex 1 Real Roots
   5. If \(D_c\) is Complex, following subtypes of Roots are possible
      1. 3 Distinct Complex Roots
      2. 1 Distinct Complex 2 Conjugate Complex Roots
      3. 2 Distinct Real 1 Complex Roots
      4. 2 Distinct Complex 1 Real Roots
   You can use the Cubic Equation Roots Calculator to find the Roots/Solutions of Cubic Polynomial Equations.
   
   You can also download the file cubic-quartic.xlsx to calculate Roots/Solutions of Cubic Polynomial Equations (for Cubic Equation with Real Coefficients Only).