Answer
$f(x)=(x-1)^{2}(x+1)(x+2)$
Zeros: $-1,\ \ -2,\ \ $with multiplicity $1$, and
$1\ \ $ with multiplicity $2$.
Work Step by Step
$f(x)$ has at most $4$ real zeros, as the degree is $4$.
Rational zeros: $\displaystyle \frac{p}{q},$ in lowest terms,
$p$ must be a factor of $ 2,\qquad \pm 1,\pm 2,$
and $q$ must be a factor of $ 1,\qquad \pm 1$
$\displaystyle \frac{p}{q}=\pm 1,\pm 2$
Testing with synthetic division, try $x-1$
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 &1 &-3 &-1 &2 \\\hline
&1 &2 &-1 & -2 \\\hline
1& 2 &-1 &-2& |\ \ 0 \end{array}$
$f(x)=(x-1)(x^{3}+2x^{2}-x-2)$
Try synthetic division again, $ x-1$
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 &2 &-1 &-2 \\\hline
&1 &3 & +2 \\\hline
1 & 3 &2 & |\ \ 0 \end{array}$
$f(x)=(x-1)^{2}(x+3x+2)=$
... factors of 2 whose sum is 3 ... are 1 and 2:
$=(x-1)^{2}(x+1)(x+2)$
$f(x)=(x-1)^{2}(x+1)(x+2)$
Zeros: $-1,\ \ -2,\ \ $with multiplicity $1$, and
$1\ \ $ with multiplicity $2$.
