Answer
The zeros are
$3$, (which is given,)
$-1+i$ and
$-1-i.$
Work Step by Step
If k is a zero of f, then, by the factor theorem, (x-k) is a factor of f(x).
Synthetic division gives us a quotient q(x), with degree one less than f(x).
and we know the remainder is zero,
so $f(x)=(x-k)\cdot q(x).$
We use this first.
$k=3$
$\left\{\begin{array}{lllllll}
3 & | & 1 & -1 & -4 & -6 & \\
& & & 3 & 6 & 6 & \\
& & -- & -- & -- & -- & --\\
& & 1 & 2 & 2 & \fbox{$0$} &
\end{array}\right.$
$f(x)= (x-3)(x^{2}+2x+2)$
We can use the quadratic formula to find the zeros of the quotient.
$x=\displaystyle \frac{b\pm\sqrt{b^{2}-4ac}}{2a} =\frac{-2\pm\sqrt{4-4(1)(2)}}{2(1)}$
$=\displaystyle \frac{-2\pm\sqrt{-4}}{2}=\frac{-2\pm 2i}{2}=\frac{2(-1\pm i)}{2}$
$=-1\pm i$
The zeros are
$3$, (which is given,)
$-1+i$ and
$-1-i.$