One solution of the equation z^3 - 2z^2 + Bz - 30 is z=2-i. Find value of B

[`Jsh]

One solution of the equation z³ - 2z² + Bz - 30 is z=2-i
If B is a real number, find the value of B and the two other solutions of the equation.

This question was in my end of year complex numbers exam (ncea level 3) and its marked as incorrect and I'm not too sure where I've gone wrong.
I'll show my working and answers and if anyone could point out where the mistake was made that would be awesome!

My working on my test is as follows:

If one solution is z=-2-i then z=-2+i is another solution.

(z+2+i)(z+2-i)(z-a) = 0 (multiplying the solutions)
(z²+2z-zi+2z+4-2i+zi+2i-i²)(z-a) = 0 (Expanding)
(z²+4z+5)(z-a) = 0 (Simplifying)
(z ³ -az²+4z ²-4az+4z-4a +z-a) = 0 (Expanding)
(z³-(a-4)z²-(4a-1)z-a = 0 (Simplifying)

a+4 = 0 (Solving for a)
-a = -4
a = 4

(-4)³-2(4)²+B(4)-30 = 0
64-32+4B-30 = 0
4B = -126
B = -31.5

 

[Deleted member 4993]

One solution of the equation z³ - 2z² + Bz - 30 is z=2-i
If B is a real number, find the value of B and the two other solutions of the equation.

This question was in my end of year complex numbers exam (ncea level 3) and its marked as incorrect and I'm not too sure where I've gone wrong.
I'll show my working and answers and if anyone could point out where the mistake was made that would be awesome!

My working on my test is as follows:

If one solution is z=-2-i then z=-2+i is another solution.

(z+2+i)(z+2-i)(z-a) = 0 (multiplying the solutions) .....................Why should that be equal to zero?
(z²+2z-zi+2z+4-2i+zi+2i-i²)(z-a) = 0 (Expanding)
(z²+4z+5)(z-a) = 0 (Simplifying)
(z ³ -az²+4z ²-4az+4z-4a +z-a) = 0 (Expanding)
(z³-(a-4)z²-(4a-1)z-a = 0 (Simplifying)

a+4 = 0 (Solving for a) ................... How is that?

-a = -4
a = 4

(-4)³-2(4)²+B(4)-30 = 0
-64-32+4B-30 = 0 .......................That 64 should be negative (assuming all other work corrcet)

4B = -126
B = -31.5

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