Answer
All zeros of h:
$-4i\quad $(given),
$4i,\ -\sqrt{11}$, $\sqrt{11}$, and $-\displaystyle \frac{2}{3}.$
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 $u=-4i$ is a zero, its conjugate $v=+4i$ is also a zero.
$b=-(u+v)=-(4i-4i)=0$
$c=(4i)(-4i)=16$
A quadratic binomial with these zeros is $x^{2}+16 ,$
and, having the same factors, it is a factor of $h$.
Dividing $(3x^{5}+2x^{4}+15x^{3}+10x^{2}-528x-352)\div(x^{2}+16)$
$\small{
\left[\begin{array}{l}
\\
x^{2}+16\\
\\
\\
\\
\\
\\
\\
\\
\\
\\
\\
\\
\end{array}\right. \left.\begin{array}{llllll}
3x^{3} & +2x^{2} & -33x & -22 & & \\
\hline)\ 3x^{5} & +2x^{4} & +15x^{3} & +10x^{2} & -528x-352 & \\
3x^{5} & & +48x^{3} & & & \\
-- & -- & -- & & & \\
& 2x^{4} & -33x^{3} & +10x^{2} & -528x-352 & \\
& 2x^{4} & & +32x^{2} & & \\
& -- & -- & -- & & \\
& & -33x^{3} & -22x^{2} & -528x-352 & \\
& & -33x^{3} & & -528x & \\
& & -- & -- & -- & \\
& & & -22x^{2} & & -352 \\
& & & -22x^{2} & & -352\\
& & -- & -- & -- &
\end{array}\right] }$
$h(x)=(x^{2}+16)(3x^{3}+2x^{2}-33x-22)$
... the last factor can be factored in pairs,
$3x^{3}+2x^{2}-33x-22=x^{2}(3x+2)-11(3x+2)=(3x+2) (x^{2}-11 )$
$h(x)=(x^{2}+16)(3x+2)(x-\sqrt{11})(x+\sqrt{11})$
All zeros of h:
$-4i\quad $(given),
$4i,\ -\sqrt{11}$, $\sqrt{11}$, and $-\displaystyle \frac{2}{3}.$
