We need to find a polynomial function f(x) with degree 3 and zeros 2 and 2i, such that:
[tex]f(-1)=30[/tex]Since 2i is a zero, -2i is also a zero of this function. Thus, we have:
[tex]f(x)=a(x-2)(x-2\imaginaryI)(x+2\imaginaryI)[/tex]where a is a constant.
Expanding the expression on the right side, we obtain:
[tex]\begin{gathered} \\ f(x)=a(x-2)\lbrack x²-(2i)²\rbrack \\ \\ f(x)=a(x-2)(x²-4i²),\text{ i^^b2=-1} \\ \\ f(x)=a(x-2)(x²+4) \\ \\ f(x)=a(x³-2x²+4x-8) \end{gathered}[/tex]Now, using f(-1) = 30, we obtain:
[tex]\begin{gathered} 30=a\lbrack(-1)³-2(-1)²+4(-1)-8\rbrack \\ \\ 30=a(-1-2-4-8) \\ \\ 30=a(-15) \\ \\ a=\frac{30}{-15} \\ \\ a=-2 \end{gathered}[/tex]Therefore, the function is:
Answer
[tex]f(x)=-2x^{3}+4x^{2}-8x+16[/tex]
