Answer
All zeros of h:
$3i\quad $(given),
$-3i,\ -3$, $\displaystyle \frac{1}{2}$, and $4.$
Work Step by Step
If $u$ and $v $ are zeros of $x^{2}+bx+c ,$
then $(x-u)(x-v)=x^{2}-(u+v)x+uv$
equals $x^{2}+bx+c ,$
from where we extract: $b=-(u+v),\quad c=uv$
Since $u=3i$ is a zero, its conjugate $v=-3i$ is also a zero.
$b=-(u+v)=-(3i-3i)=0$
$c=(3i)(-3i)=9$
A quadratic binomial with these zeros is $x^{2}+9 ,$
and, having the same factors, it is a factor of $h$.
Dividing $(2x^{5}-3x^{4}-5x^{3}-15x^{2}-207x+108)\div(x^{2}+9)$
$\small{
\left[\begin{array}{l}
\\
x^{2}+9\\
\\
\\
\\
\\
\\
\\
\\
\\
\\
\\
\\
\end{array}\right. \left.\begin{array}{llllll}
2x^{3} & -3x^{2} & -23x & +12 & & \\
\hline)\ 2x^{5} & -3x^{4} & -5x^{3} & -15x^{2} & -207x+108 & \\
2x^{5} & & +18x^{3} & & & \\
-- & -- & -- & & & \\
& -3x^{4} & -23x^{3} & -15x^{2} & -207x+108 & \\
& -3x^{4} & & -27x^{2} & & \\
& -- & -- & -- & & \\
& & -23x^{3} & +12x^{2} & -207x+108 & \\
& & -23x^{3} & & -207x & \\
& & -- & -- & -- & \\
& & & 12x^{2} & \qquad +108 \\
& & & 12x^{2} & \qquad +108 \\
& & & -- & -- &
\end{array}\right] }$
$h(x)=(x^{2}+9)(2x^{3}-3x^{2}-23x+12)$
Trying synthetic division with $x-4:$
$\left.\begin{array}{l}
4 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & -3 & -23 &12 \\\hline
& 8 & 20 & -12 \\\hline
2& 5 & -3 & |\ \ 0 \end{array}$
$h(x)=(x^{2}+9)(x-4)(2x^{2}+5x-3)$
Trying synthetic division with $x+3:$
$\left.\begin{array}{l}
-3 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & 5 &-3 \\\hline
& -6 & +3 \\\hline
2 & -1 & |\ \ 0 \end{array}$
$h(x)=(x^{2}+9)(x-4)(x+3)(2x-1)$
All zeros of h:
$3i\quad $(given),
$-3i,\ -3$, $\displaystyle \frac{1}{2}$, and $4.$
