a)
d)
1) Since the zeros of that polynomial were given, then we can write it into the factored form. Note that there are 4 zeros, so we can write:
[tex]\begin{gathered} (x-x_1)(x-x_2)(x-x_3)(x-x_4)=0 \\ (x-(-2i))(x-2i)(x-3)(x-(-4))=0 \\ (x+2i))(x-2i)(x-3)(x+4))=0 \end{gathered}[/tex]2) To find out the corresponding polynomial then we can expand it by rewriting "i" as -1
[tex]\begin{gathered} (x+2i))(x-2i)(x-3)(x+4) \\ (x+2i)(x-2i)=x^2+4 \\ (x-3)(x+4)=x^2+4x-3x-12 \\ (x^2+4)(x^2+x-12) \\ x^4+x^3-8x^2+4x-48 \end{gathered}[/tex]3) Hence, the answers are
a)
d)
[tex]x^4+x^3-8x^2+4x-48[/tex]