Note that, x = 4 is a solution for the equation.
Then, you can use this value to factored the equation as follows
[tex]x^3 - 64 = (x-4)(x^2 +4x +16) = 0[/tex]
Now you need solve
[tex]x^2 +4x +16 = 0[/tex]
[tex]x = \frac{-b \pm \sqrt{b^2 -4ac} } {2a} = \frac{-4 \pm \sqrt{16 -64} } {2} = \frac{-4 \pm \sqrt{-48} } {2} = \frac{-4 \pm 4i\sqrt{3} } {2} =-2 \pm2i\sqrt{3}[/tex]
Therefore, the solutions are
[tex]4, -2+2i\sqrt{3}, -2-2i\sqrt{3}[/tex].
The real and complex solution for the equation has been 4, [tex]\rm -2\;+\;2i\sqrt{3},\;-2\;-\;2i\sqrt{3}[/tex]. Thus, option C is correct.
The solution of the polynomial equation with the real numbers has been termed as the real solutions, while the solutions with the discrete values have been the complex solutions.
The given polynomial equation has been [tex]\rm x^3\;-64\;=\;0[/tex].
The real solution for the equation has been 4, thereby x = 4.
The polynomial equation can be written as factored form as:
[tex]\rm x^3\;-64\;=\;(x\;-\;4)\;(x^2\;+\;4x\;+\;16)[/tex]
[tex]\rm (x\;-\;4)\;(x^2\;+\;4x\;+\;16)\;=\;0[/tex]
With the 4 as the factorization,
[tex]\rm x^2\;+\;4x\;+\;16\;=\;0[/tex]
The complex solution of the equation has been: [tex]\rm -2\;+\;2i\sqrt{3},\;-2\;-\;2i\sqrt{3}[/tex].
Thus, the real and complex solution for the equation has been 4, [tex]\rm -2\;+\;2i\sqrt{3},\;-2\;-\;2i\sqrt{3}[/tex]. Thus, option C is correct.
For more information about the polynomial equation, refer to the link:
https://brainly.com/question/25924599
