MHB What is the missing term in equation (E)?

[farewell]

Good evening everyone,
here is the statement:

Note the equation (E):

Z³-12z² + = 48z-128.

1.Check that 8 is the solution of (E)
2.a) determine real a, b, c such that for all z C (overall)
Z³ 12z²-48z-128 + = (z-8) (az² + bz + c).
b) Solve the equation in C (E)

For 1 I thought about putting (z-8) factor which gives:
(z-8) z (z²-12z + 48) -128√?
after I have a polynomial of degree 2 and calculating the discriminant I find z1: 6-i√12 and z2: 6 + i√12
and there I do not know how ... can you enlighten me

 

[MarkFL]

Hello and welcome to MHB, farewell! :D

It seems that equation (E) is missing a term, since if we remove the stray plus sign, we do not find that $z=8$ is a solution.

edit: This problem has also been posted at MMF by a user of the same name, and help is being given there.