Answer
zeros of f: $-1,\ 1,\ 2 $
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 $f(c)$.
(synthetic division: the last number in the last row is $f(c)$ )
-------------------
First, let's divide
$(x^{3}-2x^{2}-x+2)\div(x+1)\qquad (c=-1)$
$\begin{array}{lllllll}
\underline{-1}| & 1 & -2 & -1 & 2 & & \\
& & -1 & 3 & -2 & & \\
& -- & -- & -- & -- & -- & \\
& 1 & -3 & 2 & |\overline{ 0 } & &
\end{array}$
So, $(x+1)$ is a factor of $f(x).$
$f(x)=(x+1)(x^{2}-3x+2)$
we try to factor the trinomial by searching for two integers
whose product is $2$ and whose sum is $-3$....
We find $-1$ and $-2:\quad x^{2}-3x+2=(x-1)(x-2)$
So, f(x) has three factors,
$f(x)=(x+1)(x-1)(x-2)$,
and by (b) of the Factor Theorem,
zeros of f: $-1,\ 1,\ 2 $
