Solutions to z 3 − ( b + 6 ) z 2 + 8 b 2 z − 7 + b 2 = 0 , b ∈ R , z ∈ C

Question

Solutions to       z    3    −  (  b  +  6  )      z    2    +  8      b    2    z  −  7  +      b    2    =  0  ,  b  ∈      R    ,  z  ∈      C

margenar0g · Accepted Answer

Consider the cubic polynomial                    (1)                    P        (        z        )        =                  z                      3                          +        A                  z                      2                          +        B        z        +        C        ,            where the coefficients A,B  and C are real numbers. If we denote its roots by z1,z2 and z3, then it factors as                    (2)                                                        P              (              z              )                                                    =                                                      (                z                −                                  z                                      1                                                  )                                            (                z                −                                  z                                      2                                                  )                            (              z              −                              z                                  3                                            )                                                                                    =                                                      z                                  3                                            −                              (                                  z                                      1                                                  +                                  z                                      3                                                  +                                  z                                      2                                                  )                                            z                                  2                                            +                              (                                  z                                      1                                                                    z                                      2                                                  +                                  z                                      2                                                                    z                                      3                                                  +                                  z                                      1                                                                    z                                      3                                                  )                            z              −                              z                                  1                                                            z                                  2                                                            z                                  3                                            .                                          The constant term is  P  (  0  )  =  C  =  −      z          1            z          2            z          3        .In the present case   A  =  −  (  b  +  6  ),   B  =  8      b          2       and   C  =  −  7  +      b          2      . Since       z          1        =  1  +  i is a given solution, then       z          2        =            z      ¯              1        =            1      +      i        ¯    =  1  −  i is another solution, as you concluded. We thus have       z          1            z          2        =      (    1    +    i    )        (    1    −    i    )    =  2 and                    (3)                    −        7        +                  b                      2                          =        −        2                  z                      3                          ,            whose solution is                    (4)                              z                      3                          =                              7            −                          b                              2                                              2                .            Since   P  (      z    1    )  =  P  (      z    2    )  =  0, we have                    (5)                                                                                                                  (                  1                  +                  i                  )                                                  3                                            −              (              b              +              6              )                                                (                  1                  +                  i                  )                                                  2                                            +              8                              b                                  2                                                            (                1                +                i                )                            −              7              +                              b                                  2                                                                                                                  =                                      −              9              +              9                              b                                  2                                            +              i                              (                −                10                −                2                b                +                8                                  b                                      2                                                  )                            =              0              ,                                                              (6)                                                                                                                  (                  1                  −                  i                  )                                                  3                                            −              (              b              +              6              )                                                (                  1                  −                  i                  )                                                  2                                            +              8                              b                                  2                                                            (                1                −                i                )                            −              7              +                              b                                  2                                                                                                                  =                                      −              9              +              9                              b                                  2                                            +              i                              (                10                +                2                b                −                8                                  b                                      2                                                  )                            =              0              ,                                          which means that b satisfies the system                    (7)                              {                                                    −                9                +                9                                  b                                      2                                                  =                0                                                                    10                +                2                b                −                8                                  b                                      2                                                  =                0.                                                                  The solution of (7) is b=-1. Using (4) we find                    (8)                              z                      3                          =        3.

Solutions to z^3−(b+6)z^2+8b^2z−7+b^2=0, b in R, z in CC

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