According to the Conjugate Roots Theorem, the complex zeroes of a polynomial always occur in pairs with its conjugate. This means that is a+ib is the root of the equation then its conjugate a-ib must also be a root of the same polynomial.
Given that x=2+2i is a root, then x=2-2i will also be a root of the given cubic polynomial.
Note that since the degree of polynomial is 3, there can be maximum 3 zeroes. Two of these are known complex zeroes, there is only one zero remaining i.e. not occurring in pair. So it must be a real number.
Let the third root be 'a', this means (x-a) ia a factor of the polynomial.
[tex]\begin{gathered} (x-2-2i)(x-2+2i)(x-a)=x^3-8x+32 \\ ((x-2)^2-(2i)^2)(x-a)=x^3-8x+32 \\ (x^2-4x+4+4)(x-a)=x^3-8x+32 \\ (x^2-4x+8)(x-a)=x^3-8x+32 \\ (x-a)=\frac{x^3-8x+32}{x^2-4x+8} \end{gathered}[/tex]
Apply the Long Division,
It is found that,
[tex]\begin{gathered} x-a=x-4 \\ a=4 \end{gathered}[/tex]
Thus, the third root of the given cubic polynomial is 4.
