since they all have a multiplicity of 1, that simply means they're all there only once, not twice or thrice or more, just once.
now, we have a root of "i", which using a + bi notation, we can simply write as 0 + i.
bearing in mind "complex roots" never come all by their lonesome, their sister is always with them, namely their conjugate is there too, so if 0 + i is here, her sister 0 - i is also a root too.
which means, there are four roots, so the equation is a quartic one, or 4th degree polynomial.
let's also recall that i² = -1.
[tex]\bf \begin{cases}
x=-4\implies &x+4=0\\
x=0+i\implies &x-i=0\\
x=0-i\implies &x+i=0\\
x=2\implies &x-2=0
\end{cases}
\\\\\\
(x+4)\stackrel{\textit{difference of squares}}{(x-i)(x+i)}(x-2)=\stackrel{y}{0}
\\\\\\
(x+4)\boxed{(x^2-i^2)}(x-2)=0\implies (x+4)\boxed{(x^2-(-1))}(x-2)=0
\\\\\\
(x+4)\boxed{(x^2+1)}(x-2)=0\implies (x+4)(x^3-2x^2+x-2)=0
\\\\\\
x^4-2x^3+x^2-2x+4x^3-8x^2+4x-8=0
\\\\\\
x^4+2x^3-7x^2+2x-8=0[/tex]
now, since it has a leading coefficient of 3, we simply multiply the whole thing by 3, since usually in the equation to 0, that factor is negligible.
[tex]\bf 3(x^4+2x^3-7x^2+2x-8)=0\implies 3x^4+6x^3-21x^2+6x-24=\stackrel{0}{y}[/tex]
