Answer
All zeros of h:
$3-2i\quad $(given),
$3+2i,\ -2$, and $5.$
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 $u=3-2i$ is a zero, its conjugate $v=3+2i$ is also a zero.
$b=-(u+v)=-(3-2i+3+2i)=-6$
$c=(3-2i)(3+2i)=9+4=13$
A quadratic binomial with these zeros is $x^{2}-6x+13 ,$
and, having the same factors, it is a factor of h.
Dividing $(x^{4}-9x^{3}+21x^{2}+21x-130)\div(x^{2}-6x+13)$
$\left[\begin{array}{l}
\\
x^{2}-6x+13\\
\\
\\
\\
\\
\\
\\
\\
\\
\\
\end{array}\right. \left.\begin{array}{lllll}
x^{2} & -3x & -10 & & \\
\hline)\ x^{4} & -9x^{3} & +21x^{2} & +21x-130 & \\
x^{2} & -6x^{3} & +13x^{2} & & \\
-- & -- & -- & & \\
& -3x^{3} & +8x^{2} & +21x-130 & \\
& -3x^{3} & +18x^{2} & -39x & \\
& -- & -- & -- & \\
& & -10x^{2} & +60x-130 & \\
& & -10x^{2} & +60x-130 & \\
& & -- & -- & \\
& & & 0 &
\end{array}\right.$
$h(x)=(x^{2}-6x+13)(x^{2}-3x-10)$
... two factors of -10 whose sum is -3 are -5 and +2 ...
$h(x)=(x^{2}-6x+13)(x-5)(x+2)$
All zeros of h:
$3-2i\quad $(given),
$3+2i,\ -2$, and $5.$
