Answer
All zeros of f:
$1+3i\quad $(given),
$1-3i,\ -1$, and $6.$
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 $u=1+3i$ is a zero, its conjugate $v=1-3i$ is also a zero.
$b=-(u+v)=-(1+3i+1-3i)=-2$
$c=(1+3i)(1-3i)=1+9=10$
A quadratic binomial with these zeros is $x^{2}-2x+10 ,$
and, having the same factors, it is a factor of f.
Dividing $(x^{4}-7x^{3}+14x^{2}-38x-60)\div(x^{2}-2x+10)$
${\small \left[\begin{array}{l}
\\
x^{2}-2x+10\\
\\
\\
\\
\\
\\
\\
\\
\\
\\
\end{array}\right. \left.\begin{array}{lllll}
x^{2} & -5x & -6 & & \\
\hline)\ x^{4} & -7x^{3} & +14x^{2} & -38x-60 & \\
x^{4} & -2x^{3} & +10x^{2} & & \\
-- & -- & -- & & \\
& -5x^{3} & +4x^{2} & -38x-60 & \\
& -5x^{3} & +10x^{2} & -50x & \\
& -- & -- & -- & \\
& & -6x^{2} & +12x-60 & \\
& & -6x^{2} & +12x-60 & \\
& & -- & -- & \\
& & & 0 &
\end{array}\right]}$
$h(x)=(x^{2}-2x+10)(x^{2}-5x-6)$
... two factors of $-6$ whose sum is $-5$ are $-6$ and $+1$ ...
$h(x)=(x^{2}-2x+10)(x-6)(x+1)$
All zeros of f:
$1+3i\quad $(given),
$1-3i,\ -1$, and $6.$
