In all cases, if [tex]f[/tex] has real coefficients, then any complex roots occur in conjugate pairs, so if [tex]a+bi[/tex] is a root, then so is [tex]a-bi[/tex]. Also, by the fundamental theorem of algebra, if [tex]r_1,\ldots,r_n[/tex] are roots to [tex]f[/tex], then for some constant [tex]a\in\mathbb R[/tex],
[tex]f(x)=a(x-r_1)\cdots(x-r_n)[/tex]
1. If [tex]n=3[/tex] and [tex]f(3)=f(2i)=0[/tex], then
[tex]f(x)=a(x-3)(x-2i)(x+2i)=ax^3-3ax^2+4ax-12a[/tex]
Given that [tex]f(-1)=50[/tex], we have
[tex]f(-1)=a(-1-3)(-1-2i)(-1+2i)=-20a=50\implies a=-\dfrac52[/tex]
[tex]\implies\boxed{f(x)=-\dfrac52x^3+\dfrac{15}2x^2-10x+30}[/tex]
2.
[tex]f(x)=a(x-4)(x-(-5+2i))(x-(-5-2i))=a x^3 + 6 a x^2 - 11 a x - 116 a[/tex]
With [tex]f(2)=-636[/tex], we have
[tex]f(2)=a(2-4)(2+5-2i)(2+5+2i)=-106a=-636\implies a=6[/tex]
[tex]\implies\boxed{f(x)=6x^3+36x^2-66x-696}[/tex]
The rest are done in the same exact way.
