Answer:
[tex]\displaystyle f(x) = -2(x-2)(x^2+4)[/tex]
Step-by-step explanation:
We want to find a third degree polynomial with zeros x = 2 and x = 2i and f(-1) = 30.
First, note that by the Complex Root Theorem, since x = 2i is a root, x = -2i must also be a root.
Hence, we will have the three factors:
[tex]\displaystyle f(x) = a(x-(2))(x-(2i))(x-(-2i))[/tex]
Where a is the leading coefficient.
Expand and simplify the second and third factors:
[tex]\displaystyle \begin{aligned} (x-(2i))(x-(-2i)) &= (x-2i)(x+2i) \\ \\ &= x(x-2i)+2i(x-2i) \\ \\ &= (x^2 - 2ix) + (2ix - 4i^2) \\ \\ &=x^2 + 4\end{aligned}[/tex]
Hence:
[tex]\displaystyle f(x) = a(x-2)(x^2+4)[/tex]
Since f(-1) = 30:
[tex]\displaystyle \begin{aligned} f(x) &= a(x-2)(x^2+4) \\ \\ (30) &= a((-1)-2)((-1)^2+4) \\ \\ 30 &= -15a \\ \\ a&= -2\end{aligned}[/tex]
In conclusion, third degree polynomial function is:
[tex]\displaystyle f(x) = -2(x-2)(x^2+4)[/tex]
