#### Answer

zeros of f: $-1,\ 2,\ 3 $

#### Work Step by Step

We will use The Factor Theorem
Let $f(x)$ be a polynomial.
$\mathrm{a}$. If $f(c)=0$, then $x-c$ is a factor of $f(x)$.
$\mathrm{b}$. If $x-c$ is a factor of $f(x)$, then $f(c)=0\quad $(c is a zero of f).
Also, The Remainder Theorem
If the polynomial $f(x)$ is divided by $x-c$, then the remainder is $f(c)$.
(in synthetic division, the last number in the last row is $f(c)$ )
-------------------
First, let's divide
$(x^{3}-4x^{2}+x+6)\div(x+1)\qquad (c=-1)$
$\begin{array}{lllllll}
\underline{-1}| & 1 & -4 & 1 & 6 & & \\
& & -1 & 5 & -6 & & \\
& -- & -- & -- & -- & -- & \\
& 1 & -5 & 6 & |\overline{ 0 } & &
\end{array}$
So, $(x+1)$ is a factor of $f(x).$
$f(x)=(x+1)(x^{2}-5x+6)$
we try to factor the trinomial by searching for two integers
whose product is $6$ and whose sum is $-5$....
We find $-2$ and $-3:\quad x^{2}-5x+6=(x-2)(x-3)$
So, f(x) has three factors,
$f(x)=(x+1)(x-2)(x-3)$,
and by (b) of the Factor Theorem,
zeros of f: $-1,\ 2,\ 3 $